Solid Thermodynamics Essay

Solid Thermodynamics Essay
SOLID MECHANICS
DYNAMICS
PRE–REQUISITE STUDIES TUTORIAL 1
LINEAR AND ANGULAR DISPLACEMENT, VELOCITY AND
ACCELERATION
This tutorial is essential for anyone studying the group of tutorials on beams.
Essential pre–requisite knowledge for Edexcel HNC Mechanical
Principles UNIT 21722P.
Essential pre–requisite knowledge for the Engineering Council Diploma
Exams D209 Mechanics of Solids and D225 Dynamics of Mechanical
Systems.
Essential pre–requisite knowledge for the Engineering Council
Certificate Exam C105 Mechanical and Structural engineering.
Covers part of the syllabus for the Engineering Council Certificate Exam
C103 Engineering Science.
On completion of this tutorial you should be able to
Define linear motion.
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Figure 3
Average velocity = total distance travelled/time taken
Average velocity = area under graph/base length
Since the base length is also the time taken it follows that the area under the graph is the distance
travelled. This is true what ever the shape of the graph. When working out the areas, the true scales
on the graphs axis are used.
WORKED EXAMPLE No.2
Find the average velocity and distance travelled for the journey depicted on the graph above. Also
find the acceleration over the first part of the journey.
SOLUTION
Total Area under graph = A + B + C
Area A = 5x7/2 = 17.5 (Triangle)
Area B = 5 x 12 = 60 (Rectangle)
Area C = 5x4/2 = 10 (Triangle)
Total area = 17.5 + 60 + 10 = 87.5
The units resulting for the area are m/s x s = m
Distance travelled = 87.5 m

Time taken = 23 s
Average velocity = 87.5/23 = 3.8 m/s
Acceleration over part A = change in velocity/time taken = 5/7 = 0.714 m/s2.
© D.J.DUNN
5
SELF ASSESSMENT EXERCISE No.2
1.
A vehicle travelling at 1.5 m/s suddenly accelerates uniformly to 5 m/s in 30 seconds. Calculate the
acceleration, the average velocity and distance travelled.
(Answers 0.117 m/s2, 3.25 m/s and 97.5 m)
2.
A train travelling at 60 km/h decelerates uniformly to rest at a rate of 2 m/s2.
Calculate the time and distance taken to stop.
(Answers 8.33 s and 69.44 m)
3.
A shell fired in a gun accelerates in the barrel over a length of 1.5 m to the exit
velocity
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Essay about Chapter 1 Lab
Name _ ___________________ Motion in 2D Simulation
Go to http://phet.colorado.edu/simulations/sims.php?sim=Motion_in_2D and click on Run Now.
1) Once the simulation opens, click on 'Show Both' for Velocity and Acceleration at the top of the
page. Now click and drag the red ball around the screen. Make 3 observations about the blue and
green arrows (also called vectors) as you drag the ball around.
1. The green vector moves in the direction of the mouse until the red ball catches up to it.
2. The blue and green vectors move in opposite directions as the red ball slows down.
3. The faster you move the mouse, the longer the blue vector becomes.
2) Which color vector (arrow) represents velocity and which one represents ... Show more content
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5) What happens to the vectors when you jerk the ball rapidly back and forth across the screen?
Explain why this happens.
Both vectors lengthen when the ball is jerked quickly across the screen. This occurs because of the
speed at which it is performed causes the acceleration to increase. The velocity vector also lengthens
due to the large distance being put between the red ball and the destination at a fast rate.
6) Now click on 'Circular' on the bottom. Describe the motion of the ball and the behavior of the two
vectors. Is there a force on the ball? How can you tell? Be detailed in your explanations.
The ball is moving with a steady speed in a circle across the screen. The blue acceleration vector
points towards the center of the circle while the green velocity vector follows the edge of the circle
throughout the movement. Yes, there is a force on the ball. This is known because both acceleration
and velocity vectors are visible and the ball is moving at a constant speed and in a pattern.
7) Click on 'Simple Harmonic' on the bottom. Based on the behavior of the ball and the vectors,
write a definition of Simple Harmonic Motion.
The simple harmonic motion shows that both velocity and acceleration vectors move in harmony
when the speed of the ball is increasing in a linear direction. When speed is decreasing, the vectors
become

Physics of Skydiving Essay
The Physics of Skydiving
What Is Skydiving?
Skydiving is an adrenaline–based sport with a fairly simple concept –– jump from a high place
(usually out of a plane) from several thousand feet above sea level and hope and pray for a safe
landing. This safe landing is often times achieved through the use of a device called a parachute,
which enables the skydiver to reduce his speed to such a point that colliding with the earth will not
be fatal.
This paper will explain a few of the key concepts behind the physics of skydiving. First we will
explore why a skydiver accelerates after he leaps out of the plane before his jump, second we will
try and explain the drag forces effecting the skydiver, and lastly we will attempt to explain how ...
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Lets examine a free body diagram of this situation.
One may now clearly see why Joe does not accelerate downward while he remains is in the plane.
The normal force of the plane is acting against the normal force of the gravity acting on Joe's 100
kilogram mass (this is getting close to the maximum weight allowed for skydiving). Since the sum
of the forces in the y, or upward direction, is equal to zero, there is effectively no force acting on
Joe. Thus, the acceleration of Joe must be zero.
Now that Joe understands why he is remaining at a constant altitude, he's ready to take his first
jump. Joe is given the signal to jump from the jump master, and he steps right off the edge of the
plane. Joe is now instantly accelerating in the downward direction. Joe's acceleration will soon cause
him to travel at a rapid rate. But exactly how fast will he be traveling at a certain time? Without this
knowledge, Joe could easily splatter against the surface of the earth. In order to calculate Joe's
velocity at any time, T, we use the following equation:
X = Vi * T + .5 * g * t^2
Where X is the change in distance, Vi is the initial velocity, t is time, and g is the acceleration due to
gravity. Since we know Joe's initial velocity was 0, and the acceleration of gravity is always 9.8
m/s^2, we can calculate how far Joe has traveled at any point in time. Lets examine the new free
body diagram for this situation.

As you can see, the normal force is now

The Importance Of Walking Under Steady State Conditions
The current knowledge available from the literature on walking at different speeds assumes a direct
relationship between the level of activity of lower extremities muscles and speed. Several studies in
the past have looked into this relationship for walking under steady–state conditions. In this section
we review these studies and their findings and we highlight limitations of their approaches and
discuss the suggestions to expand the analysis to walking under transient conditions.
Hof et al. [21] were one of the first groups to measure surface electromyograms (EMGs) from
subjects walking at different speeds. Five speeds were instructed for over ground walking and
EMGs were measured for ten walking steps at each speed. Parts of EMG profiles showed
considerable changes with speed while other parts did not show any changes. The authors suggested
that the activity at any speed could be estimated through a linear interpolation of sum of the profile
at the lowest speed and a function that represents the increase in the activity per unit increase of
normalized speed. One limitation of the calculations here was lack of statistical tests. Because of
this shortcoming the linear interpolation model could be used to make reliable conclusions only
about the effect of speed on the activity in the vicinity of the peak of each signal. Without building
the confidence intervals we cannot infer the effect of speed on activation profiles over the entire gait
cycle. In addition to lack of

Lab Report On Newton 's Second Law
Makenzie Tobin and Anna Zytchkova
Period: 5
Lab Report
November 24, 2015
Title:
Introduction: In this lab we were trying to prove right Newton 's Second Law which is F=ma. We
had used mass and force to test if the law was would match up with our experiment. In the first
experiment we used a car and put a certain amount of weights in it to see how much it would affect
the mass, time and velocity. In the second experiment, we stretched a rubber band from 1cm to 5cm
and released the car from the length of the rubber bands to calculate the time and velocity. In both
problems to calculate velocity we used, 1 cm divided by the time calculated by the photogate. Our
hypothesis was that if more mass is added, then acceleration will decrease and if force increases
then acceleration will also increase. In both experiments we had calculated the velocity. Since
acceleration is the rate of which velocity either increases or decreases, we could see if the car was
accelerating or decelerating in the experiments.
Methods:
Materials:
–Car –3 weights –Photogate –Rubber band – Regular 12 inch ruler (using centimeters side)
Procedures:
1) These are the steps used to test how mass affects acceleration: –First you push the car through the
photogate with no weights in it. Next you push the car with only a single weight in it to push it
through the photogate. Then you add a second weight so

Physics Lab Report
ABSTRACT:
The lab of one dimensional motion is a series of experiments that deal with different types of motion
in a single direction. In the first experiment, one dimensional motion of a small cart on an air track
is measured in a one photogate system. The acceleration was calculated by the infrared light
emitting electrode of the photogate sensing the slacks on the picket fence. The calculation for
gravity yielded 9.63 m/s^2, which is consistent with the accepted value of 9.8m/s^2. In the second
experiment, acceleration of a cart traveling down a slight incline was measured with a two photogate
method. Gravity was calculated and yielded a value of 8.7429, a bit lower than the accepted value of
9.8 m/s^2. The third experiment was ... Show more content on Helpwriting.net ...
The second length of time will be the time in between when the first photogate is blocked until when
the second photogate is unblocked (deltaT12). The third time measured will be the time that the
second photogate is blocked (deltaT2). The average velocity that it takes for the flag to pass through
each photogate will be calculated by the change in the known distance of the flag divided by the
time it takes to pass through the photogate in question. The acceleration of the cart can then be
calculated by equation 1.2.
A = V2–V1 / DeltaT12 Equation 1.2
The next experiment will be the measurement of acceleration due to a horizontal force, measured by
a single photogate. The force will come from a weight hanging vertically from a string that runs
through a pulley that is connected horizontally to the cart, measure by a single photogate.
Figure 1.3:Horizontal Force cause by Hanging Weight The acceleration can be expressed by
considering three free body diagrams. First the sum of the Y forces on the hanging weight. Ó F(y) =
mg – T1 = ma Equation 1.3
Notice that T1 is equal to mg – ma. Next, evaluate the sum of the horizontal forces acting on the

Projectile Lab Report
Projectile 2 Lab Report
Kelsy Ecclesiastre and Sara Lopez
AP Physics 1
Period 6
Objective/Purpose:
To correctly identify the range of a projectile when fired above the horizontal axis with only a meter
stick (i.e. no timer).
Background:
According to Henderson, speed is how far an object travels during a specific amount of time and
does not include the direction. Velocity can be explained as the speed of an object, but also including
the direction the object is heading towards (Henderson, 2015).
Giancoli states that the velocity of an object is taken with the magnitude– the numerical value– and
the direction of the object (Giancoli, 2009). Therefore, with these two qualities, velocity can be
illustrated as a vector.
Acceleration occurs when ... Show more content on Helpwriting.net ...
After performing more than five trials, choose the five closest points on the paper and find the
average of those five points. At least ten trials could be performed if the lab was being repeated.
References:
Elert, G. (2015). Acceleration. Retrieved August 25, 2015, from http://physics.info/acceleration/
Giancoli, D. (2009). Physics: Principles with Applications. Upper Saddle River, NJ: Pearson
Education.
Heald, J. (2015). What is Acceleration? – Definition and Formula. Retrieved August 23, 2015, from
http://study.com/academy/lesson/what–is–acceleration–definition–and–formula.html
Henderson, T. (2015). Speed and Velocity. Retrieved September 6, 2015, from
http://www.physicsclassroom.com/class/1DKin/Lesson–1/Speed–and–Velocity
Lucas, J. (2014, June 26). Force, Mass & Acceleration: Newton's Second Law of Motion. Retrieved
August 25, 2015 from

Vector
VECTOR FUNCTIONS
VECTOR FUNCTIONS
Motion in Space: Velocity and Acceleration
In this section, we will learn about:
The motion of an object using tangent and normal vectors.
MOTION IN SPACE: VELOCITY AND ACCELERATION
Here, we show how the ideas of tangent and normal vectors and curvature can be
used in physics to study:
 The motion of an object, including its velocity and acceleration, along a space curve.
VELOCITY AND ACCELERATION
In particular, we follow in the footsteps of
Newton by using these methods to derive
Kepler's First Law of planetary motion.
VELOCITY
Suppose a particle moves through space so that its position vector at
time t is r(t).
VELOCITY
Vector 1
Notice from the figure ... Show more content on Helpwriting.net ...
 So, C=i–j+ k

VELOCITY & ACCELERATION
Example 3
It follows: v(t) = 2t2 i + 3t2 j + t k + i – j + k
= (2t2 + 1) i + (3t2 – 1) j + (t + 1) k
Example 3
Since v(t) = r'(t), we have: r(t) = ∫ v(t) dt
= ∫ [(2t2 + 1) i + (3t2 – 1) j + (t + 1) k] dt = (⅔t3 + t) i + (t3 – t) j + (½t2 + t) k + D
Example 3
Putting t = 0, we find that D = r(0) = i.
So, the position at time t is given by: r(t) = (⅔t3 + t + 1) i + (t3 – t) j + (½t2 + t) k
The expression for r(t) that we obtained
in Example 3 was used to plot the path of the particle here for 0 ≤ t ≤ 3.
In general, vector integrals allow us to recover:
 Velocity, when acceleration is known
v(t )
v(t0 )
t t0
a(u ) du
 Position, when velocity is known
r (t ) r (t0 )

t t0
v(u ) du
If the force that acts on a particle is known, then the acceleration can be found from
Newton's Second Law of Motion.
The vector version of this law states that if,
at any time t, a force F(t) acts on an object of mass m producing an acceleration a(t),
then
F(t) = ma(t)
Example 4
An object with mass m that moves in
a circular path with

Essay On Trikonasana
MOTION ANALYSIS OF TRIKONASANA
ABSTRACT: The objective of this study is to analyse human motion during Trikonasana from a
kinematic and inverse dynamic human model and computational biomechanical solver,
LifeMOD[1]. Low cost Inertial sensors(IMU) were used to obtain segment parameters for
Trikonasana. Trikonasana is one of the important postures(asana) practised in yoga. Trikonasana has
to be performed with perfection to obtain maximum benefits, which can be quantified by its
symmetry, grace and consistency[2]. This study helps to determine torque, angular velocity and
angular acceleration at hip joint and shoulder joint using a mathematical model. These kinematic
parameters obtained from the mathematical model for Trikonasana are validated from the inverse ...
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The kinematic model has three degrees of freedom in segment angles ፀ . Both the arms are assumed
to be at the same angle as the postures involve symmetry of the arms in trikonasana . But the model
can be easily extended to include independent arm motions. The three segments are assumed to be
rigid bodies with their mass concentrated at their center of mass. The different segment lengths,
masses, center of mass locations and moment of inertia are obtained from anthropometric data. The
Joints possess rotational stiffness and frictional damping. Lagrange's method was used to
mathematically model the posture. The equation of motion are derived using the Lagrangian
equation. ddt(∂L∂θi)–∂L∂θi=Qi where i =1,2,3... where Lagrangian L= K.E – P.E
Qi includes the generalized forces and torques on the system

An Experiment On Proportional Navigation
Proportional Navigation For short–to–medium range homing missiles the most widely used law in
practice is proportional navigation (PN) guidance.
The missile acceleration should nullify the line–of–sight (LOS) rate between the target and
interceptor that is basic philosophy behind PN. Originally, PNG law creates angular velocity or
acceleration commands perpendicular to the LOS (line of sight). If two bodies are closing on each
other eventually they will intercept when there is no rotation in the line of sight (LOS) between the
two bodies relative to the inertial space.
By making the interceptor missile heading proportional to the LOS rate for a non–maneuvering
target the PN guidance law seeks to null out the LOS rate. More specifically, the PN guidance law
seeks to null the LOS rate against non–maneuvering targets by making the interceptor missile
heading proportional to the LOS rate. For instance, in flying a proportional navigation course, the
missile attempts to null out any line–of–sight rate that may be developing. The missile does this by
commanding wing deflections to the control surfaces. Consequently, these deflections cause the
missile to execute accelerations normal to its instantaneous velocity vector. Thus, the missile
commands g's to null out measured LOS rate. As will be developed in the discussion that follows,
this relation can be expressed as follows:
V ̇_PNG=NV_λ λ ̇ (3.1)
Where
V ̇_PNG: The

Vector Rk4 ( Const Vector
Vector RK4(const Vector &X, const Vector &Xdot, double t, double h, int resting, int stopped){
Vector K1(4), K2(4), K3(4), K4(4);
K1 = h * Xdot; K2 = h * dynamics(X + 0.5 * K1, t + 0.5 * h, resting, stopped); K3 = h *
dynamics(X + 0.5 * K2, t + 0.5 * h, resting, stopped); K4 = h * dynamics(X + K3, t + h, resting,
stopped);
return X + (K1 + 2 * K2 + 2 * K3 + K4) / 6.0;
}
Vector Euler(const Vector &X, const Vector &Xdot, double h){
return X + h * Xdot;
}
void XToxv(Vector3d &x, Vector3d &p, Vector3d &l, Quaternion &q, const Vector &X){ x.x =
X[0]; x.y = X[1]; x.z = X[2]; p.x = X[3]; p.y = X[4]; p.x = X[5]; l.x = X[6]; l.y = X[7]; l.z = X[8];
q.w = X[9]; q.x = X[10]; q.y = X[11]; q.z = X[12];
} ... Show more content on Helpwriting.net ...
Given current state and time, return rate of change of state
//
Vector dynamics(const Vector &X, double t, int resting, int stopped){ Vector3d x, p, l; Quaternion q;
Vector newXdot(13); Vector3d F; double fn, ft, vt; double friction;
// compute total spring and gravity force XToxv(x, p, l, q, X); F = pl.m * g – sp.k * ((x – x0).norm()
– sp.l0) * (x – x0).normalize();
// if in resting contact compute the friction force and support force fn = F * pl.n; // force component
normal to the surface if(resting && fn < 0){
if(stopped){ // static friction ft = F * pl.tangent; // force component tangent to the surface friction =
–Min(Abs(pl.mus * fn), Abs(ft)) * Sign(ft); } else{ // dynamic friction vt = v * pl.tangent; // velocity
component tangent to the surface friction = pl.mud * fn * Sign(vt); }
// add the friction force F = F + friction * pl.tangent;

// subtract support force normal to the resting surface F = F – fn * pl.n; }
// build the state rate

Kinematics
Worksheet Kinematics 1. A body starts from rest and reaches a speed of 5 m/s after travelling with
uniform acceleration in a straight line for 2 s. Calculate the acceleration of the body.
2. A body starts from rest and moves with uniform acceleration of 2m/s2 in a straight line.
a. Calculate the velocity after 5s.
b. Calculate the distance travelled in 5s.
c. Find the time taken for the body to reach 100m from its starting point.
3. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground.
Determine the distance traveled before takeoff.
4. A ... Show more content on Helpwriting.net ...
Determine the skidding distance of the car (assume uniform acceleration). 12. A kangaroo is capable
of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo. 13. If Michael
Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to
move upwards to the peak and then return to the ground)? 14. A bullet leaves a rifle with a muzzle
velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of
0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration). 15. A baseball is
popped straight up into the air and has a hang–time of 6.25 s. Determine the height to which the ball
rises before it reaches its peak. (Hint: the time to rise to the peak is one–half the total hang–time.)
16. The observation deck of tall skyscraper 370 m above the street. Determine the time required for
a penny to free fall from the deck to the street below. 17. A bullet is moving at a speed of 367 m/s
when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m.
Determine the acceleration of the bullet while moving into the clay. (Assume a uniform
acceleration.) 18. A stone is dropped into a deep well and is heard to hit the water 3.41 s after being
dropped. Determine the depth of the well. 19. It was once recorded that a Jaguar left skid marks

Lab Report On Projectile Motion
The purpose of this lab was to examine projectile motion and how the horizontal motion is
independent of the vertical motion. Data was used to calculate gravity given that the vertical
acceleration was only due to g. The data was collected by sending a ball down a ramp that had a
horizontal end. At the end of the ramp there was a monitor that recorded the velocity of the ball as
left the ramp. Carbon paper was laid on the ground where the ball landed; the spot that was left
behind was measured from the origin that was directly below the end of the ramp. The height of the
ramp was also measured, which was 0.757 m. The two balls diameter and mass were measured. Ball
one had the diameter of .0255 m and the mass of 0.0601 kg, ball two had the diameter of 0.0242 m
and the mass of 0.0209 kg.
Sample Calculation, for trial 1, Ball 2:
Flight time:
Vertical acceleration:
Rearranged kinematic equation to solve for acceleration given that the initial velocity is equal to
zero:
Ball 1 had an average vertical acceleration of –12 with a standard deviation of 0.20. For ball 2, it
had an average vertical acceleration of –11 with a standard deviation of 0.32 . The expected value
for the vertical acceleration was –9.80 . Using that, the percent difference for ball 1 was 18% and for
ball 2 it was 8.6%. Both average vertical accelerations were larger in magnitude then the expected
value; they were not very accurate. Though ball 2 was more accurate than ball 1 because it had a
lower

Physics Problems
UNIFORMLY ACCELERATED MOTION Problem–Solving
1. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car
(assume uniform acceleration). 2. A bike accelerates uniformly from rest to a speed of 7.10 m/s over
a distance of 35.4 m. Determine the acceleration of the bike. 3. An engineer is designing the runway
for an airport. Of the planes which will use the airport, the lowest acceleration rate is likely to be 3
m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is
the minimum allowed length for the runway? 4. If basketball player has a vertical leap of 1.29 m,
then what is his takeoff speed and his hang time (total time to move ... Show more content on
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Hammy has jumped with initial velocity of 5 m/s at an angle of 250 for 1 second. Will he be able to
reach the ice block? If not maintaining the same velocity and hang time, what should be the angle of
his take off to be able to reach the ice block?
2. The target is located 100 m away from the cannon. The target is aimed at 250 angle. The target
was hit after 2s. What is the velocity of the cannon ball upon being fired? At what distance from the
ground is the peak of the projectile?
3. A bullet is fired with a velocity of 200 m/s at an angle of 450 from the horizontal, after
approximately 2s, the target is hit. How far is the target placed?
B. Target is higher than Launch point
1. The horizontal distance of the cannon from the target is 100 m away. However the target is on top
of a hill with a height of 15 m. The cannon is aimed at 500 angle from the horizontal and upon firing
cannon ball has a velocity of 150 m/s. Will the target be hit by the cannon? 2. Your opponent is
located at a horizontal distance of 5m. But he is elevated at a height of 3m. Your cannon always fire
the cannon ball at a velocity of 120m/s, the hang time of the cannon ball is 1.5 s. What should be
your aim angle to be able to hit your opponent? With that aim angle, how high is the peak of the
projectile? 3. You are located uphill at a height of 6m from the ground. Your opponent is at a
horizontal distance of 260 m away from

Designing A Rubber Band Powered Car Essay
Introduction The objective of this project was to construct a rubber band–powered car. This process
had two aspects: the assembly of the car and the analysis of the car. With trial and error, the car was
assembled successfully. Throughout the project, my partner and I used our knowledge of one–
dimensional kinematics. Completing the project with another person utilized both our collaboration
and communication skills. During the analysis process, we assessed the velocity and acceleration of
the car. The rubber band–powered car had to meet a set of specifications. The car needed to possess
three wheels, travel at least three meters in a straight line, have constant acceleration, move
dependably and easily, and look professional. Over the course of this project, my understanding of
one–dimensional kinematics increased by the hands–on creation of a rubber band–powered car.
Materials and Methods The rubber band powered–car was assembled with materials provided in the
classroom and supplemented with materials my partner and I had in our homes. The final product
has a strong cardboard base with a tube in the middle to thread the rubber band through. Connected
with hot glue, thin plastic dowels are attached to the wheels. Lightweight wheels lined with duct
tape are used to move the car. Bottle caps stabilize the CD wheels and dowels. A small plastic nub is
glued to one dowel. The rubber band is knotted onto the back dowel, stretched through the tube, and
placed over or under

The 5th wave
1. A 1.08 x 103 kg car uniformly accelerates for 12.0s from rest. During this time the car travels 132
m north. What is the net force acting on the car during this acceleration 2. A net force of 12 N is
exerted on and to cause it to accelerate at a rate of 0.03 m/s2. Determine the mass of the
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MERGEFORMATINET 5. 4. How much force does a 40,000kg rocket develop toaccelerate 2m/s2
5. Block 1 is 12kg and block 2 is 23kg calculate ... Show more content on Helpwriting.net ...
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Physics : An Analysis Of The Physics Of Physics
The diagram can be simplified as F1 is equivalent to FF and FG combined.
Using these diagrams, it is easier to calculate an approximate value for both the parallel component
of the normal force and the maximum force of the eddy currents. The mass of the cart m remains
constant at 507.85g ± .01g, and the peak acceleration due to eddy currents in the 90 cm run was –.7
ms–2 ± .1 ms–2. These values can be inputted into Newton's Second Law:
(F_1+F_E )=(.50785)*(–.7)
To calculate F1, the acceleration of the cart before the aluminum strip is needed, which is found to
be .6 ms–2 ± .1 ms–2 in all trials. The mass of the cart remains the same,
F_1=(.50785)*(.6)
F_1= .3 N
The percentage uncertainty for F1 can be calculated by adding the percentage ... Show more content
The data provides evidence for an equilibrium state too, where the force of the eddy currents
eventually equals the force of the gravity parallel to the track. This was seen in the lower initial
velocity runs, but would be expected for the higher runs as well, if the track and aluminum strip
were to continue. These results are significant, particularly for those who might want to use eddy
currents to help stop an object, such as rollercoasters. It demonstrates that eddy currents are most
effective when the object is fast, and that other means are necessary to totally stop an object.
Moreover, those seeking to implement eddy currents as a braking system will want to know what
variables affect the strength of the braking force, and these experiments serve to answer that. Error
Analysis and Future Improvements
A source of error within Experiment 1 was with measuring the time taken. Though better than a
stopwatch and human reaction, there was still difficulty in being accurate. Using the slow–motion
analysis software, it was possible to get increments of .01 seconds. However, it wasn't clear exactly
when the casing was dropped, hence the uncertainty of ± .04 seconds. In addition, the total mass was
not always 74.00 g ± .01g, but instead varied from 73.20 g ± .01g to 75.13 g ± .01g, because pennies
were used as additional mass. This may have influenced the time, and so in the future it would be
better to use a more

Ph1004 Lab1 Report
PH1004 Experiment 1
One Dimensional Motion
Name: Jin Cai
Student ID: 0392789
Partner: Kai Lan
Section: G1
Date Experiment Performed: Jan 29th
Date Experiment Due: Feb 5th
Introduction:
(This should be in paragraph form and should explain the objective and the related theory. For
example, what is the overall objective of the experiment, and what are you going to measure to
accomplish it? The Introduction should not exceed one page.)
The objective of the experiment is to understand the meaning of displacement, average velocity,
instantaneous velocity, average acceleration and instantaneous in one–dimensional motion. The first
experiment was to calculate the average and instantaneous velocity. We first ... Show more content
Dev. |2.510 |
Best Estimate Value for aav: 40.894 (cm/s2)
Uncertainty for aav: 2.510 (cm/s2)
Final Results
(For this experiment, the final results are your values for vav, v, and aav, with their uncertainties.

Make sure to round properly, and to present the fractional uncertainties as well)
[pic]0.10/31.16 = 0.32%
[pic]0/0.86=0%
[pic]2.51/40.89=6.14%
The best estimate values for the average and instantaneous velocity and average acceleration are
31.207 (cm/s), 0.858 (cm/s) and 40.894 (cm/s2).
Conclusion
(This is your chance to show what you have learned, and it should be written in paragraph

Unit 2 Physics Question Paper
1. Infer whether the size of an object's displacement (amount of units between start and stop) could
be greater than the distance the object travels. The displacement of an object cannot be greater than
the distance an object travels because traveling from point A to point B equals displacement. The
displacement can't be greater but the distance can. 2. Describe the motion represented by a
horizontal line on a distance–time graph. A horizontal line on a distance–time graph means the
object being graphed is not moving at all. 3. Explain whether, during a trip, a car's instantaneous
speed (movement of an object at a specific instant) can ever be greater than its average speed. Yes, a
car's instantaneous speed can be greater than its average

Physics 1 Midterm Exam Review 2
PHYSICS 1 MIDTERM EXAM REVIEW #2
1. A 6.0–N force and an 8.0–N force act concurrently on a point. As the angle between these forces
increases from 0° to 90°, the magnitude of their resultant (1) decreases (2) increases (3) remains the
same
2. A car increases its speed from 9.6 m/s to 11.2 m/s in 4.0 s. The average acceleration of the car
during this 4.0–second interval is (1) 0.40 m/s2 (3) 2.8 m/s2 (2) 2.4 m/s2 (4) 5.2 m/s2
3. What is the speed of a 2.5–kilogram mass after it has fallen freely from rest through a distance of
12 m? (1) 4.8 m/s (3) 30. m/s (2) 15 m/s (4) 43 m/s
4. A machine launches a tennis ball at an angle of 25° above the horizontal at a speed of 14 m/s. The
ball returns to level ground. ... Show more content on Helpwriting.net ...
Compared to the horizontal component of the car's velocity at point A, the horizontal component of
the car's velocity at point B is (1) smaller (2) greater (3) the same
18. A force of 25 N east and a force of 25 N west act concurrently on a 5.0–kg cart. What is the
acceleration of the cart? (1) 1.0 m/s2 west (3) 5.0 m/s2 east (2) 0.20 m/s2 east (4) 0 m/s2
19. Centripetal force Fc acts on a car going around a curve. If the speed of the car were twice as
great, the magnitude of the centripetal force necessary to keep the car moving in the same path
would be (1) Fc (2) 2 Fc (3) Fc / 2 (4) 4 Fc
20. A 1.0 x 105–newton truck at rest on a hill that makes an angle of 8.0° with the horizontal. What
is the component of the truck's weight parallel to the hill? (1) 1.4 x 103 N (3) 1.4 x 104 N (2) 1.0 x
104 N (4) 9.9 x 104 N
21. A book of mass m falls freely from rest to the floor from the top of a desk of height h. What is
the speed of the book upon striking the floor? (1) (2gh)½ (3) mgh (2) 2gh (4) mh
22. A 0.149–kg baseball, initially moving at
15 m/s, is brought to rest in 0.040 s by a baseball glove on a catcher's hand. The magnitude of the
average force exerted on the ball by the glove is (1) 2.2 N (3) 17 N (2) 2.9 N (4) 56 N

23. Which body is in equilibrium? (1) a satellite moving around Earth in a circular orbit (2) a cart
rolling down a frictionless

Phys216 Lab3
Worksheet for Interactive Lab3: Circular Motion
Name _____________ Physics 216 Grade ___/42 Date ______
Circular motion is one of the fundamental building blocks for understanding the physical world
because it is so common. This interactive lets you examine the basic ideas of kinematics and forces
for objects moving in circular motion.
1. (6 points) Use the controls under Show Vectors to draw arrows for several different quantities.
The velocity vector is drawn in a special way: a big, bold arrow showing its direction and
magnitude, plus smaller arrows for its horizontal and vertical components. * How are the directions
of the acceleration and force vectors related?
Force has a magnitude and a direction, and it is a vector. ... Show more content on Helpwriting.net
...
A) Always to the left.
B) Always to the right.
C) Always to upward.
D) Always to downward.
E) Always away from the center of the circular path.
F) Always toward the center of the circular path.
Question 2 (3 points)
In what direction does the velocity of the mass point?
A) Always to the left.
B) Always to the right.
C) Always to upward.
D) Always to downward.
E) Always away from the center of the circular path.
F) Always toward the center of the circular path.
G) Always perpendicular to the force.
In what direction does the acceleration of the mass point? A) Always away from the center of the
circular path.
B) Always toward the center of the circular path.
C) Always perpendicular to the force.

D) Always in the opposite direction of the force.
What would happen to the force if I increase the velocity? A) It will get larger.
B) It will get smaller.
C) It will not be affected.
Angular velocity is not described by ...? A) Revolutions per minute
B) Meters per second
C) Radians per minute
D) Angular displacement
An object is in uniform circular motion. Identify the correct statements.
1. Its acceleration is constant (acceleration is a vector quantity). 2. Its radial acceleration component
is constant in magnitude. 3. Its

Formula Booklet Physics 11th
Call 1600–111–533 (toll–free) for info.
Formula Booklet – Physics XI
Dear students Most students tend to take it easy after the board examinations of Class X. The
summer vacations immediately after Class X are a great opportunity for the students to race ahead of
other students in the competitive world of IITJEE, where less than 2% students get selected every
year for the prestigious institutes. Some students get governed completely by the emphasis laid by
the teachers of the school in which they are studying. Since, the objective of the teachers in the
schools rarely is to equip the student with the techniques reqired to crack IITJEE, most of the
students tend to take it easy in Class XI. Class XI does not even have the pressure of ... Show more
content on Helpwriting.net ...
Ph: 55270275, 55278916 E–16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585
4
Call 1600–111–533 (toll–free) for info.
(1)
Formula Booklet – Physics XI
(a) (b) (c)
F12 = − F12 m F = F21 M+m F Acceleration: a = M+m
→
→
M (2) m F21 Fig. 1
F FR 2 = ; m I = mG g GM
→ →
Contact force: F12 =

F12
F
(xvi) Inertial mass: mI = F/a (xvii) Gravitational mass: mG =
→
(xviii) Non inertial frame: If a 0 be the acceleration of frame, then pseudo force F = − m a 0
Example:

Trebuchet Arm Ratio Hypothesis
Arm ratio
It was hypothesised for this variable that the longer the arm length, the greater the distance will be
because of the centripetal force formula:
Fc=mv2r
When rearranged,
Fc rm=v
Shows that a higher radius value will result in a higher final velocity.
Results obtained from testing showed that the longer the arm ratio, the further the horizontal
displacement will be.
During testing, videos were taken of the trebuchet in action. This was used to calculate the launch
velocity and various angles. The calculations for this, shown in Appendix 1, are somewhat reliable
but may be inaccurate in some areas.
The theoretical calculation for output force from the trebuchet arm was calculated (Appendix 3) and
the result was compared to the result from the ... Show more content on Helpwriting.net ...
In contrast to this hypothesis, it was discovered that the experimental values deviate from the
theoretical values when the counterweight falls more than 20 cm. This is due to the trebuchet
shifting and rocking during launch which meant significant energy was lost during the launch.
The height of the counterweight was chosen as a variable because of the formula for GPE
(gravitational potential energy) is GPE=MGHwhich is all multiplication, meaning the greater the
height, the greater the GPE value will be. A higher GPE means higher energy input into the system
which, according to the law of conservation of energy, means that theoretically, the output energy
that ends up in the projectile is also greater. The law of conservation of energy which states that
energy can be neither created nor destroyed, instead, it is transformed from one type to another.
The range likely started to drop after the release height of 22cm because this is the point at which
the trebuchet shifts too much when launched which impacts the effectiveness of the

Mechanics
UNIVERSITI TUNKU ABDUL RAHMAN
|Centre |: Centre for Foundation Studies (CFS) |Unit Code |: FHSC1014 |
|Course |: Foundation in Science |Unit Title |: Mechanics |
|Year/ Trimester |: Year 1 / Trimester 1 |Lecturer | |
|Session |: 201605 | | |
Additional Tutorial 2: Vector and translational kinematics.
1. Find the x and y–components of: (a) a displacement of 200 ... Show more content on
Helpwriting.net ...
Tutorial 2: Vector and translational kinematics.
1. (a) Dx = 200 cos 30.0o = 173 km Dy = 200 sin 30.0o = 100 km (b) vx = 40.0 cos 120o = –20.0
km/h vy = 40.0 sin 120o = +34.6 km/h (c) Fx = 50.0 cos 330o = 43.3 N; Fy = 50.0 sin 330o = –25.0
N
2. [pic]
| Vector |x–component |y–component |
|F1 |–(21.0) sin30.0o = –10.5 N |+(21.0) cos30.0o = +18.2 N |
|F2 |+15.0 N |0 N |
|F3 |–(–10.5+15.0) = –4.5 N |–(18.2+0) = –18.2 N |
[pic] [pic]
3. Wx = 200 sin 300; Wx = 100 N, down the plane. Wy = 200 cos 300; Wy = 173 N, normal to the
plane.
4. The triple jump consists of a double jump in one direction, followed by a perpendicular single
jump, which we can represent with displacement vectors J and K and D = J + K. J = 4d and K = 2d
where [pic]. [pic] [pic]
5. (a) [pic] (b) When the faster car has a 15.0 min lead, it is ahead by a distance equal to that
traveled by the slower car in a time of 15.0 min. [pic] The faster car pulls ahead of the slower car at
a rate of

Review Questions For Chap 2 1 Essay
Remedial Physics––––––––––––––––––––––––––––––––––––Review Questions
–––––––––––––––––––––––––––––––––––––Chapter 2
Quantitative Problems
1) An object moves 15.0 m north and then 11.0 m south. Find both the distance traveled and the
magnitude of the displacement vector.
A) 6.0 m, 26.0 m B) 26.0 m, 4.0 m C) 26.0 m, 26.0 m D) 4.0 m, 4.0 m
2) A boat can move at 30 km/h in still water. How long will it take to move 12 km upstream in a
river flowing 6.0 km/h?
A) 20 min B) 22 min C) 24 min D) 30 min
3) 55 mi/h is how many m/s? (1 mi = 1609 m.)
A) 25 m/s B) 49 m/s C) 90 m/s ... Show more content on Helpwriting.net ...
It then starts to snow and you slow to 35 mi/h. You arrive home after driving 4 hours and 15
minutes. How far is your hometown from school?
A) 180 mi B) 190 mi C) 200 mi D) 210 mi
11) A motorist travels 160 km at 80 km/h and 160 km at 100 km/h. What is the average speed of the
motorist for this trip?
A) 84 km/h B) 89 km/h C) 90 km/h D) 91 km/h
12) An airplane travels at 300 mi/h south for 2.00 h and then at 250 mi/h north for 750 miles. What
is the average speed for the trip?
A) 260 mi/h B) 270 mi/h C) 275 mi/h D) 280 mi/h
13) An airplane increases its speed from 100 m/s to 160 m/s, at the average rate of 15 m/s2. How
much time does it take for the complete increase in speed?
A) 17 s B) 0.058 s C) 4.0 s D) 0.25 s
14) A car traveling 60 km/h accelerates at the rate of 2.0 m/s2. How much time is required for the
car to reach a speed of 90 km/h?
A) 15 s B) 30 s C) 45 s D) 4.2 s
15) A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintain that velocity for 10 s,

and then decelerates at the rate of 2.0 m/s2 for 4.0 s. What is the final speed of the car?
A) 20 m/s B) 16 m/s C) 12 m/s D) 10 m/s
16) A cart with

Velocity and Smart Timer
Cruz, Romer Kevin C. Oct. 29, 2013
2010150921 Nov. 05, 2013
PHY10L/A11
Experiment # 2
KINEMATICS
Abstract – Kinematics of linear motion is defined as the studies of motion of objects without
considering the effects that produce the motion. This experiment will show how to determine the
linear motion with constant (uniform) velocity particularly the dynamic cart and linear motion with
constant (uniform) acceleration, (e.g. free fall of motion). At the end of the experiment we found out
that the velocity is a speed that involves direction of an object as well as the time. While for the
acceleration, it is directly proportional to the distance or height but inversely proportional to the
time. By close ... Show more content on Helpwriting.net ...
Adjust the track if the cart moves until it became stationary. We then attached the picket fence to the
cart. The first photo gate was set mounting at 25 cm mark and the second at 65 cm mark to the track.
We adjust the photo gates height so the picket fence attached to the cart will not touch the photo
gate. We securely placed the gates to the track perpendicular to minimize errors.
The phone plug of both photo gates are connected to the smart timer, photo gate 1 to channel 1 and
photo gate 2 to channel 2. We then set the mode of the smart timer to measure TIME, TWO GATES.
And press the third button on the smart timer to start/restart. The cart is placed at 0 cm mark on the
track. The cart was launched be pressing the trigger, we take the reading on the smart timer and
record this as the time interval for the first trial. The displacement is figured by subtracting the
distance of the first photo gate to the second photo gate. The cart is launch 5 times more, with each
launch adjusting the position of the photo gate 2 by 10 cm. Then we compute for the cart's average
speed. After the data gathering, we plot a graph with displacement against time.
Part B: Determination of Acceleration Due to Gravity Using Cart's Acceleration.
We set up the track with the 0 cm end elevated using a stand, with the end stop at the lower end and
record it as final position, Xf. The height, H of the track is adjusted to 10 cm as its initial height. We
placed the photo

Report on Motion Essay
Table Of Contents
PHS 100–552 Lab
Part I: Scenario H Graph...................................................... 2
Scenario H Regions and Force Diagrams..................................3
Region and Force Diagram Information....................................4
Part II: Graph 6 ...................................................................5
Step–By–Step Instruction........................................................6
Regions and Force Diagrams...................................................7
Region Information................................................................8
Newton's Laws..................................................................... 9
Self–Assessment...................................................................10
Scenario H
You are stopped at a stop sign. Your friend pushes your car forward at an increasing velocity for two
seconds. She then pushes your car for three more seconds at a constant velocity. Your friend stops
pushing and you immediately apply the brakes for ... Show more content on Helpwriting.net ...
Velocity and Acceleration graphs become negative due to the slowing of the cart.
The cart is stopped at 2.0 meters. Net force equals zero. All three graphs show the cart is stopped
with a flat line across 2.0 meters.
Newton's Laws
Region A
Region B
Region C
Region D
Region E

Newton's 1st Law: The cart is at rest and will remain at rest until a force is applied.
Newton's 2nd Law: The two forces acting upon the cart (hand and fan) are equal so there is no
acceleration. At 2.5 seconds the acceleration is changed because of the force of hand.
Newton's 3rd Law: The force of hand is applied at 2.5 seconds so the cart moves towards the sensor
as a reaction to the force.
Newton's 1st Law: The force of hand is applied which puts the cart in motion towards the sensor.
Newton's 2nd Law: The force of hand is applied to the cart which changes the Acceleration and
moves the cart towards the sensor.
Newton's 3rd Law: Force of hand was applied at 2.5 seconds in Region A, the force of hand built
momentum until the cart moves at the start of Region B. This motion continued until the force of fan
overcame the force of hand, forcing the cart way from the sensor.
Newton's 1st Law: Force of fan is applied to the cart which overcomes the force of hand putting the
cart in motion away from the sensor.
Newton's 2nd Law: Force of fan is applied changing the direction of the cart away from the sensor.

Physics
Experiment
Picket Fence Free Fall
5
We say an object is in free fall when the only force acting on it is the earth's gravitational force.
No other forces can be acting; in particular, air resistance must be either absent or so small as to be
ignored. When the object in free fall is near the surface of the earth, the gravitational force on it is
nearly constant. As a result, an object in free fall accelerates downward at a constant rate.
This acceleration is usually represented with the symbol g.
Physics students measure the acceleration due to gravity using a wide variety of timing methods.
In this experiment, you will have the advantage of using a very precise timer connected to the
calculator and a Photogate. The ... Show more content on Helpwriting.net ...
If the acceleration of your Picket Fence appears constant, fit a straight line to your data.
a.
b.
c.
d.
e.
5–2
Press ENTER , and select RETURN TO MAIN SCREEN from the SELECT GRAPH screen.
Select ANALYZE from the main screen.
Select CURVE FIT from the ANALYZE screen.
Select LINEAR (VELOCITY VS TIME) from the SELECT CURVE FIT screen.
Record the slope of the fitted line in the Data Table.
Modified from and reported with permission of the publisher Copyright (2000),
Vernier Software & Technology
Physics with Calculators
Picket Fence Free Fall
f. Press ENTER to see the fitted line with your data.
g. To return to the main screen, press ENTER , and then select RETURN TO ANALYZE SCREEN,

finally selecting RETURN TO MAIN SCREEN.
9. To establish the reliability of your slope measurement, repeat Steps 6 through 9 five more times.
Do not use drops in which the Picket Fence hits or misses the Photogate. Record the slope values in
the Data Table.
DATA TABLE
Trial
1
2
3
4
5
6
2
Slope (m/s )
Minimum
Maximum
Average
2
Acceleration (m/s )
Acceleration due to gravity, g
Precision
±
m/s2
%
ANALYSIS
1. From your six trials, determine the minimum, maximum, and average values for the acceleration
of the Picket Fence. Record them in the Data Table.
2. Describe in words the shape of the distance vs. time graph for the free

The Performance Of The Car Under Test
The analysis of the performance of the car under test was completed using AVL Cruise software.
Various Graphs are obtained so that the actual performance of the car can be determined. Graph 1
(Abundance) prominently shows the torque of the engine at the respective engine speed along with
the specific fuel consumption at that speed. Approximately 230Nm is the maximum torque that can
be obtained at 4200 rpm, trailed by decrease in torque for the remaining speed. A cycle run was
conducted for three different periods of cycle in which the vehicle was accelerated then hold and
decelerated for certain time depending upon the requirement of cycle. Graph 2–A shows the change
in velocity upon change in acceleration for three different cycle times. Also the shift of gears during
the period can be observed. Graph 2–B illustrates consumption of the fuel during the three different
stages. It can be noted that when the velocity was about 50 km/h the fuel consumption was about 5.0
l/h. Also it significantly notifies that during idling the fuel consumption is 1.6 l/h. In this Graph it
can be inferred that the maximum operating pressure is 12 Bar for a speed of 4200rpm. Maximum
fuel is consumed during the start of the engine at 0.5 bar, fuel consumption gradually decreases with
increase in engine speed. Graph 3 speaks about the maximum operating pressure is 12 Bar for a
speed of 4200rpm. Maximum fuel is consumed during the start of the engine at 0.5 bar and also
between 0 bar and 1.6 bar at

Determining G on an Incline Essay
Determining g on an Incline
Lab #1
Theory: During the early part of the seventeenth century, Galileo experimentally examined the
concept of acceleration. One of his goals was to measure the acceleration due to gravity, or the
acceleration of freely falling objects. Unfortunately, his timing devices were not precise enough to
measure the free fall time directly. He decided to "dilute" gravity by using fluids, inclined planes,
and pendulums.
Galileo's idea of diluting gravity using inclined planes worked like this: the acceleration of a rolling
cart on an inclined plane is small, therefore is easy to measure; when the angle of the incline gets
bigger, the acceleration will get bigger; by measuring the dependence of the acceleration on ... Show
more content on Helpwriting.net ...
Using Vernier, we clicked collect while releasing the cart after motion detector starts to click. This
was done moving the hand quickly out the path. Using logger pro, indicated which portion was to be
used by dragging across the graph to indicate the starting and ending times. Then the linear button
was clicked to perform the linear regression of the selected data. The Linear Button was used to
determine the slope of the velocity vs. time graph, only using the portion of the data for times when
the cart was freely rolling. We found the acceleration of the cart from the fitted line. Record the
value in the data table. These steps where repeated 5 mores times. Measured the length of the
incline, x which is the distance between the two points of the ramp. Measure the height, h, the height
of the book(s). The last two measurements was used determine the angle of the incline. Raise the
incline by placing a second book under the end. Adjust the book so that distance, x, is the same as
the previous reading. Repeated these steps with 3, 4 and 5 books.
ANALYSIS DATA
The greater the incline and greater the height, the greater the acceleration of the cart. –This
experiment measured the acceleration of a cart moving down a sloped track in order to find a
relation between the acceleration of an object and the sine of the angle at which it is moving. There
were three separate experimental stages, with the variable being the angle at which the

How Height Affect The Speed Of A Toy Car
The hypothesis was "If the height of the ramp increases then the speed of the toy car will increase
because the the time taken to get all the way down will increase, so speed will increase." The data
supports the hypothesis. As the height of the ramp increased so did the speed of the car. This shows
a positive trend. When the ramp was 10 cm high, after rolling for 30 cm, its speed was 23 cm/s.
When the ramp was 15 cm high, after rolling for 30 cm the speed was 33.3 cm/s. At 10 cm, when
the car rolled for 60 cm, its speed was 26.9 cm/s. On the contrary, at 15 cm high when the car rolled
for 60 cm it travelled at an average rate of 38.5 cm/s. Finally, at 10 cm high, after rolling 90 cm, the
toy car was travelling at and average rate of 35 cm/s. On the contrary, the car that started on the
ramp 15 cm high, after rolling for 90 cm, was travelling 50.3 cm/s. The data clearly shows a positive
trend because when the height of ... Show more content on Helpwriting.net ...
As the height of the ramp increased so did the acceleration of the car. This shows a positive trend.
When the ramp was 10 cm high, between 0 and 30 cm, the car accelerated at an average rate of 17.7
cm/s2. When the ramp was 15 cm high, between 0 and 30 cm the car accelerated at a rate of 37
cm/s2. At 10 cm high, between 30 and 60 cm the car accelerated at a rate of 4.2 cm/s2 . At 15 cm
high, between 30 and 60 cm, the toy car accelerated at an average rate of 7.9 cm/s2 . Lastly, at 10
cm, the car accelerated at a rate of 23.8 cm/s2 between 60 and 90 cm. When the ramp was 15 cm
high, between 60 and 90 cm the car accelerated ata rate of cm/s2 . The data clearly shows a positive
trend because when the height of the ramp increased so did the acceleration of the car. This may be
because the car took less time to get down the ramp and had more

Rationale And Issues That Occurs When Managers Read A...
This paper will discuss the rationale and issues that occurs when managers read a popular book by a
management guru and hastily tries to implement its idea and one–size–fits–all recommendations
without proper regards for the organization's unique problems and needs. The quick–fix mentality
that fosters this problem can be avoided by remaining current with management literature from the
field, ensure that the concepts are applied based on science rather than advocacy. "The First 90
Days: Proven Strategies for Getting Up to Speed Faster and Smarter" by Michael D. Watkins is a
remarkable book that covers all the issues listed above with real scenario. Mr. Watkins was asked by
Johnson & Johnson's Corporate Management Development Group (J & J) to develop workshops and
coaching processes to accelerate the company's leaders in transition, and his work soon evolved into
an engaging partnership with J&J as the testing bed for his new development and deployment of his
ideas. I really enjoyed reading this book, because it made me think of all the times when I faced
these problems and did not understand why or what went wrong in many situations by reading the
book it changed the way I think about many things. The actions are taken during the first few
months in a new role will largely determine if there is success or failure, when this happens the
problems can be traced to vicious cycles that develop in the first few months on the job, whether
failure to lose your opportunities or

Physics Acceleration
Chapter 6. Uniform Acceleration
Problems:
Speed and Velocity
6–1. A car travels a distance of 86 km at an average speed of 8 m/s. How many hours were required
for the trip?
[pic] [pic] t = 2.99 h
6–2. Sound travels at an average speed of 340 m/s. Lightning from a distant thundercloud is seen
almost immediately. If the sound of thunder reaches the ear 3 s later, how far away is the storm?
[pic] t = 58.8 ms
6–3. A small rocket leaves its pad and travels a distance of 40 m vertically upward before returning
to the earth five seconds after it was launched. What was the average velocity for the trip?
[pic] v = 16.0 m/s
6–4. A car travels along a U–shaped curve for a distance of 400 m in 30 ... Show more content on
Helpwriting.net ...
For Problem 6–17, what is the maximum displacement from the bottom and what is the velocity 4 s
after leaving the bottom? (Maximum displacement occurs when vf = 0)
2as = vo2 – vf2; [pic]; s = +21.3 m vf = vo + at = 16 m/s = (–6 m/s2)(4 s); vf = – 8.00 m/s, down
plane 6–19. A monorail train traveling at 80 km/h must be stopped in a distance of 40 m. What
average acceleration is required and what is the stopping time? ( vo = 80 km/h = 22.2 m/s)
2as = vo2 – vf2; [pic]; a = –6.17 m/s2
[pic] ; t = 3.60 m/s
Gravity and Free–Falling Bodies
6–20. A ball is dropped from rest and falls for 5 s. What are its position and velocity? s = vot +
½at2; s = (0)(5 s) + ½(–9.8 m/s2)(5 s)2 ; s = –122.5 m vf = vo + at = 0 + (–9.8 m/s2)(5 s); v = –49.0
m/s
6–21. A rock is dropped from rest. When will its displacement be 18 m below the point of release?
What is its velocity at that time? s = vot + ½at2; (–18 m) = (0)(t) + ½(–9.8 m/s2)t2 ; t = 1.92 s vf =
vo + at = 0 + (–9.8 m/s2)(1.92 s); vf = –18.8 m/s
6–22. A woman drops a weight from the top of a bridge while a friend below measures the time to
strike the water below. What is the height of the bridge if the time is 3 s? s = vot + ½at2 = (0) + ½(–
9.8 m/s2)(3 s)2; s = –44.1 m

6–23. A brick is given an initial downward velocity of 6 m/s. What is its final velocity after falling a
distance of 40 m?
2as = vo2 – vf2 ; [pic]; v = (28.6 m/s;

Essay on The Moment of Inertia of a Disk and a Ring
Objective: The objective of this laboratory was to theoretically calculate the moment of inertia of a
disk and a ring and then to verify the moment of inertia for both objects through experiment. This
laboratory shows that while the theoretical is not within the uncertainty of the experimental, both
values are extremely similar to each other.
Data and Analysis:
Data:
Table 1: The Angular Acceleration of No Ring and Ring
Trial No Ring Ring
5g 4.57 ± 0.005 rad/s2 1.32 ± 0.005 rad/s2
10g 13.16 ± 0.005 rad/s2 3.09 ± 0.005 rad/s2
15g 20.45 ± 0.005 rad/s2 4.83 ± 0.005 rad/s2
20g 27.89 ± 0.005 rad/s2 6.60 ± 0.005 rad/s2
25g 35.65 ± 0.005 rad/s2 8.35 ± 0.005 rad/s2 Table 2: The Average Experimental and Theoretical
Moment of Inertia for No ... Show more content on Helpwriting.net ...
This calculation will be done later on. After using equation 1 to find the experimental moment of
inertia, the average and standard deviation of the five trials for No Ring and Ring were calculated
using Excel commands for average and standard deviation. The averages will be used later on in
order to calculate the experimental moment of inertia for the ring. In the next step, the theoretical
moment of inertia was calculated for the disk by using equation 2.
I_disk^th=1/2 M_d R_d^2 (2)
In the equation above, the I_disk^th is the theoretical moment of inertia of the disk, M_d is the mass
of the disk, Rd is the radius of the disk. Next, the theoretical moment of inertia of the Ring was
calculated using equation 3 below.
I_ring^th=1/2 M_r (R_1^2+R_2^2) (3)
In the equation above, I_ring^this the theoretical moment of inertia of the Ring, Mr is the mass of
the ring, R_1^2 is the inner radius of the ring, and R_2^2 is the outer radius of the ring. In order to
compare the experimental and theoretical moments of inertia of the Ring, the experimental moment
of inertia of the ring alone has to be calculated. It is given that the experimental moment of the
system equals to the experimental moment of inertia of the ring plus the experimental moment of
inertia of the disk. The experimental moment of inertia can be found by solving

Motion Down an Incline Essay
Lab 1, problem 3: Motion Down an Incline
Shaoren Yuan
October 5, 2013
Physics 1301W, professor: Dr. Zudov, TA: David
Abstract
The processes of a cart rolling up and returning back along a track were recorded, and the processes
(motion of the cart.) were described as equations. Also, we calculated the accelerations of every
stage (aup, adown and ahighest). Then the relationship among aup, adown and ahighest was
concluded. Finally, the acceleration was measured and was proved from data.
Introduction
If there is a car launched from the bottom of an incline and it goes up until reaching the highest
point, then it reverses its direction. To ensure the safety under this circumstance, the accelerations of
every stage need to be ... Show more content on Helpwriting.net ...
Therefore, the net force of the block was mg*sinθ and pointed down along the track.
The free–body diagram is as following:
Since no friction acted on cart, aup=adown=ahighest=–g*sinθ.
Procedure:
The thickness of two blocks together (H) was measured, and the length of the track (L) was
recorded. Then the two wood blocks were put on the desk, and one end of the track was set on the
blocks to make the incline. Meanwhile, the angle between the track and the desk, θ, was calculated
(technically, sinθ was measured according to sinθ=H/L). Next, we marked 30cm on the one side of
the track by using tapes, which was applied to determine the distance in the Motion
Lab&VideoRECOREDER. After that, the distance and height of the camera was adjusted to a proper
position to make sure record the motion well. According to the video effect from computer, the best
position for observation was about 40cm height above the desk, and 80cm away from the track.
Then we put the cart with the compressed spring on the lower end of the track. After that, we fixed
the track in case that the track would move or vibrate when the cart started. Then the spring of the
cart was released and the time was recorded simultaneously. After the cart returned back, before it
collided the end of the track, we then stopped it and finished recording. Next, the data

Does Wheel It Go Farther?
"Wheel" it Go Farther?
Data Analysis
The question that was asked and answered from this experiment was if the size of the wheel affects
the distance traveled. The hypothesis was that if the spool size is larger, then the spool will travel
farther. The independent variable is the spool, while the dependent variable is the distance traveled.
The control variables were the ramp height, the place the experiment took place, and the material of
the ramp. The procedures for this experiment were to: Get 3 spools– a large, medium, and small. Get
a ramp and elevate it 5 centimeters off of the ground. The spools shall roll down the ramp three
times each, in roughly the same spot.
In the experiment, there were three spools. Each spool rolled down the ramp three times, in roughly
the same spot each time. The small spool had an average distance of 1.11 meters. The medium–sized
spool rolled an average of 2.40 meters. The largest spool traveled an average of 3.01 meters.
Patterns that were seen in the data were that the small spool had a smaller average than the medium
spool and the large spool, the medium spool had a larger average than the small spool but a smaller
average than the large spool, and the large spool had a larger average than the medium and small
spool. The larger the spool, the farther distance it traveled. ... Show more content on Helpwriting.net
...
There is a relationship between the variables. The independent variable relates to the dependent
variable because wheels come in all different shapes and sizes, and are put on various types of
vehicles. Something that can be learned from this data is that the larger the spool, the farther
distance it will travel. A possible reason that the larger spool traveled the farthest is that the spool
had more momentum. The data that was collected accepted my hypothesis, and that the larger the
spool, the farther it will

Inclined Plane Experiment
The purpose of this experiment is to test an Independent variable by collecting experimental data
and its corresponding Dependent data. In this experiment it was tested what happens to a dynamic
trolley or skateboard as different experimental variables are applied. Those experimental variables
were different weights. Which means that this experiment is about the physics principles of the laws
of motion and the inclined plane. The laws of motion are called newtons laws. There are three laws.
These laws are an object at rest tends to stay at rest and an object in motion tends to stay in motion
unless acted upon by an unbalanced force. This is the law of inertia. The second law is F=ma and the
third law is every action has an equal or opposite ... Show more content on Helpwriting.net ...
The last factor is acceleration which definitely also effects how fast the object moves down the
inclined plain. The acceleration is the gravitational pull of 9.8 m/s/s important factors in this
experiment is friction and air resistance between the car and the inclined plain. Friction is involved
as when the object slide down ramps, friction is involved, and the force of friction opposes the
motion down the ramp. (Steven Holzner, 2014) The force of friction is proportional to the force
from the ramp that balances the component of gravity that is perpendicular to the ramp. Air
resistance also is an important part of this experiment as air resistance and friction are directly
involved in the movement of the object and can slow down the process to convert gravitational
potential energy into kinetic energy. ("BBC – GCSE Bite size: Gravitational potential energy
(GPE)", 2016) Another force involved in this experiment is the gravitational pull. This is also
another important force involved in this experiment and the movement of the car down the inclined
plain. In the case of this experiment different weights was used to test the acceleration of the heavier
vehicle to test the hypothesis and theme of the experiment. In this experiment Gravitational
Potential Energy (GPE) and Kinetic Energy (KE) both play an important

Lab Report On Projectile
What is a projectile? In "what is a projectile" (Physics Classroom, 2015), a projectile is a moving
object and the only force acting upon it is gravity. The actual path of a projectile could vary
according to the position and direction of the launch of the projectile. The image 'types of
projectiles' (2015) shows three types of projectiles. P1 is the projectile which has the movement
only in vertical direction while P2 has more vertical and a little horizontal motion thus moving in
two dimensions. P3 has both motions almost equal. Projectile has the characteristic of being under
influence the force of gravity only. If the path of a moving object is affected significantly by some
air resistance, the object is not a projectile any more. It is also not a projectile while at the source or
when it hits something during its flight. A ball shown in the diagram below is not a projectile till the
ball leaves the hands of the person dropping it or throwing it. The air resistance can be ignored at
this point because the effect is not significant. It again seizes to be a projectile when it is in contact
with the ground because the force of the reaction from the ground and the force of friction has come
into action. P1 P2 P3 Diagram 1 Assumptions: Several examples of projectiles will be under
discussion during the following study, so it is

The Components of Newton's Laws of Motion
Velocity is the time rate of change of position of an object in a particular direction. Velocity along a
straight line is known as linear velocity and is commonly measured by meters per second (m/s).
Since both speed and direction are implemented in the measurement velocity the direction must be
given. Velocity is a vector quantity, which includes magnitude or speed and a direction into account.
An object doesn't need to move in a straight–line path to have linear velocity. Instantaneous velocity
of any point of an object undergoing circular motion is a vector quantity. When an object is forced to
follow a curved path it has instantaneous linear velocity at any point of its travel. Velocity is
calculated by dividing the time it took to travel the distance into the distance it traveled or V=d/t
(Rusk, Dr. Rogers D. (2014). Velocity. In AccessScience. McGraw–Hill Education. Retrieved from
http://www.accessscience.com/content/velocity/729500) Acceleration is the time rate of change in
velocity. Instantaneous acceleration is the limit of the rate of change in velocity to the time taken to
change velocity. When the acceleration is constant, the average acceleration and the instantaneous
acceleration are equal. When unbalanced forced act on an object, the objects will undergo
acceleration. A force is the influence on an object, which causes it to accelerate. If the object doesn't
change direction the object will have a constant acceleration. Acceleration is

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