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Fart A stant If you ignore the weight of the bar itself, how far from the left end of the barbell is the center of gravity? Express your answer to three significant figures and include the appropriate units A 1.80-m-long barbell has a 20.0 kg weight on its left end and a 36.0 kg weight on its right end You may want to review (Pages 204- 205) HA Submit Previous Answers Request Answer X Incorrect; Try Again Part B Where is the center of gravity if the 7.00 kg mass of the barbell itself is taken into account? Express your answer to three significant figures and include the appropriate units HA dValue Units Solution Part A : If we ignore the weight of the bar itself, how far from the left end of a barbell is the center of gravity? we know that, x cg = (m 1 x 1 + m 2 x 2 ) / (m 1 + m 2 ) x cg = [(20 kg) (0 m) + (35 kg) (1.8 m)] / [(20 kg) + (35 kg)] x cg = (63 kg.m) / (55 kg) x cg = 1.14 m Part B : Where is the center of gravity if the 7 kg mass of a barbell itself is taken into account? we know that, x cg = (m 1 x 1 + m 2 x 2 + m 3 x 3 ) / (m 1 + m 2 + m 3 ) x cg = [(20 kg) (0 m) + (35 kg) (1.8 m) + (7 kg) (1.8 m / 2)] / [(20 kg) + (35 kg) + (7 kg)] x cg = [(63 kg.m) + (6.3 kg.m)] / (62 kg) x cg = [(69.3 kg.m) / (62 kg)] x cg = 1.11 m .

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Lecture14

Today's physics lecture covered torque and rotational equilibrium. Key points included definitions of torque as the rotational effect of a force and rotational inertia. Examples were worked through to demonstrate torque, equilibrium conditions requiring the sum of forces and sum of torques to equal zero, and applications to problems involving objects in static equilibrium situations. An exam review and homework problems related to these concepts were also discussed.

Lecture14

The document summarizes key points from Physics 101 Lecture 14 on torque and equilibrium:
1. The lecture covers rotational kinetic energy, rotational inertia, torque, and static equilibrium conditions.
2. Torque is defined as the rotational effect of a force and is calculated as τ = rF × sinθ. Static equilibrium requires the sum of all torques and forces to equal zero.
3. Examples are provided to demonstrate calculating torque for different situations and using equilibrium conditions to solve problems involving forces and torques.

Sistemas de partículas y conservación del momento lineal

1. The center of mass of a system of three 2 kg point masses located at x=0, x=0.2 m, and x=0.5 m is located at x=0.23 m.
2. The center of mass of a boy of 24 kg located 20 m from a man of 86 kg is located 15.63 m from the boy and 4.36 m from the man.
3. The center of mass of a semicircular shell of radius R is located at (4/3)πR from the center of the circle.

Chapter 1(4)SCALAR AND VECTOR

Vectors have both magnitude and direction, represented by arrows. The sum of two vectors is obtained by placing the tail of one vector at the head of the other. If the vectors are at right angles, their dot product is zero, while their cross product is maximum. Scalar multiplication scales the magnitude but not the direction of a vector.

Mechanics of materials lecture 02 (nadim sir)

This document discusses shear force and bending moment in structural members. It defines shear force, normal force, and bending moment as the internal forces that develop in a beam due to applied loads. It presents methods for determining the shear force diagram and bending moment diagram of a beam based on the slope relationships between load, shear force, and bending moment. Several examples are worked through to demonstrate how to calculate and draw the shear force diagram and bending moment diagram for beams under different loading conditions.

Lecture04

1) Today's physics lecture covered kinematics and dynamics, including constant acceleration equations and Newton's second law.
2) An example problem involved a tractor pulling a trailer with a total acceleration of 1.5 m/s^2, finding the horizontal force on the tractor from the ground is 1050 N.
3) Another example used a pulley system with two boxes to find the common acceleration of 2.45 m/s^2 and the time for the second box to hit the table is 0.81 seconds.

Lecture04

1) Today's physics lecture covered kinematics and dynamics, including constant acceleration equations and Newton's second law.
2) An example problem involved a tractor pulling a trailer with a total acceleration of 1.5 m/s^2, finding the horizontal force on the tractor from the ground is 1050 N.
3) Another example used a pulley system with two boxes to find the common acceleration of 2.45 m/s^2 and the time for the second box to hit the table is 0.81 seconds.

Shear and moment diagram

This document discusses shear and moment diagrams for beams. It provides examples of calculating reactions, shear, and bending moment at different points along simple beams subjected to various load cases including point loads and uniform loads. Key steps include determining reactions, drawing the shear diagram, and using the shear diagram to draw the moment diagram. The maximum shear and bending moment as well as points where shear and moment equal zero are identified. Overhanging ends can cause negative bending moments. Understanding shear and moment diagrams is important for beam design.

Lecture14

Today's physics lecture covered torque and rotational equilibrium. Key points included definitions of torque as the rotational effect of a force and rotational inertia. Examples were worked through to demonstrate torque, equilibrium conditions requiring the sum of forces and sum of torques to equal zero, and applications to problems involving objects in static equilibrium situations. An exam review and homework problems related to these concepts were also discussed.

Lecture14

The document summarizes key points from Physics 101 Lecture 14 on torque and equilibrium:
1. The lecture covers rotational kinetic energy, rotational inertia, torque, and static equilibrium conditions.
2. Torque is defined as the rotational effect of a force and is calculated as τ = rF × sinθ. Static equilibrium requires the sum of all torques and forces to equal zero.
3. Examples are provided to demonstrate calculating torque for different situations and using equilibrium conditions to solve problems involving forces and torques.

Sistemas de partículas y conservación del momento lineal

1. The center of mass of a system of three 2 kg point masses located at x=0, x=0.2 m, and x=0.5 m is located at x=0.23 m.
2. The center of mass of a boy of 24 kg located 20 m from a man of 86 kg is located 15.63 m from the boy and 4.36 m from the man.
3. The center of mass of a semicircular shell of radius R is located at (4/3)πR from the center of the circle.

Chapter 1(4)SCALAR AND VECTOR

Vectors have both magnitude and direction, represented by arrows. The sum of two vectors is obtained by placing the tail of one vector at the head of the other. If the vectors are at right angles, their dot product is zero, while their cross product is maximum. Scalar multiplication scales the magnitude but not the direction of a vector.

Mechanics of materials lecture 02 (nadim sir)

This document discusses shear force and bending moment in structural members. It defines shear force, normal force, and bending moment as the internal forces that develop in a beam due to applied loads. It presents methods for determining the shear force diagram and bending moment diagram of a beam based on the slope relationships between load, shear force, and bending moment. Several examples are worked through to demonstrate how to calculate and draw the shear force diagram and bending moment diagram for beams under different loading conditions.

Lecture04

1) Today's physics lecture covered kinematics and dynamics, including constant acceleration equations and Newton's second law.
2) An example problem involved a tractor pulling a trailer with a total acceleration of 1.5 m/s^2, finding the horizontal force on the tractor from the ground is 1050 N.
3) Another example used a pulley system with two boxes to find the common acceleration of 2.45 m/s^2 and the time for the second box to hit the table is 0.81 seconds.

Shear and moment diagram

This document discusses shear and moment diagrams for beams. It provides examples of calculating reactions, shear, and bending moment at different points along simple beams subjected to various load cases including point loads and uniform loads. Key steps include determining reactions, drawing the shear diagram, and using the shear diagram to draw the moment diagram. The maximum shear and bending moment as well as points where shear and moment equal zero are identified. Overhanging ends can cause negative bending moments. Understanding shear and moment diagrams is important for beam design.

Ch01 2

To solve differential equations of the form y' = ay - b:
1) Use methods of calculus to find the general solution, which is y = Ce^(at) + b/a, where C is a constant.
2) If initial conditions are given, set them equal to the general solution to determine the value of C, yielding a unique solution.
3) The equilibrium solution, where y' = 0, is found by setting y = b/a.

From the Front LinesOur robotic equipment and its maintenanc.docx

From the Front Lines
Our robotic equipment and its maintenance represent a fixed cost of $23,320 per month. The cost-effectiveness of robotic-assisted surgery is related to patient volume: With only 10 cases, the fixed cost per case is $2,332, and with 40 cases, the fixed cost per case is $583.
Source: Alemozaffar, Chang, Kacker, Sun, DeWolf, & Wagner (2013).
MATLAB sessions: Laboratory 5
MAT 275 Laboratory 5
The Mass-Spring System
In this laboratory we will examine harmonic oscillation. We will model the motion of a mass-spring
system with differential equations.
Our objectives are as follows:
1. Determine the effect of parameters on the solutions of differential equations.
2. Determine the behavior of the mass-spring system from the graph of the solution.
3. Determine the effect of the parameters on the behavior of the mass-spring.
The primary MATLAB command used is the ode45 function.
Mass-Spring System without Damping
The motion of a mass suspended to a vertical spring can be described as follows. When the spring is
not loaded it has length ℓ0 (situation (a)). When a mass m is attached to its lower end it has length ℓ
(situation (b)). From the first principle of mechanics we then obtain
mg︸︷︷︸
downward weight force
+ −k(ℓ − ℓ0)︸ ︷︷ ︸
upward tension force
= 0. (L5.1)
The term g measures the gravitational acceleration (g ≃ 9.8m/s2 ≃ 32ft/s2). The quantity k is a spring
constant measuring its stiffness. We now pull downwards on the mass by an amount y and let the mass
go (situation (c)). We expect the mass to oscillate around the position y = 0. The second principle of
mechanics yields
mg︸︷︷︸
weight
+ −k(ℓ + y − ℓ0)︸ ︷︷ ︸
upward tension force
= m
d2(ℓ + y)
dt2︸ ︷︷ ︸
acceleration of mass
, i.e., m
d2y
dt2
+ ky = 0 (L5.2)
using (L5.1). This ODE is second-order.
(a) (b) (c) (d)
y
ℓ
ℓ0
m
k
γ
Equation (L5.2) is rewritten
d2y
dt2
+ ω20y = 0 (L5.3)
c⃝2011 Stefania Tracogna, SoMSS, ASU
MATLAB sessions: Laboratory 5
where ω20 = k/m. Equation (L5.3) models simple harmonic motion. A numerical solution with ini-
tial conditions y(0) = 0.1 meter and y′(0) = 0 (i.e., the mass is initially stretched downward 10cms
and released, see setting (c) in figure) is obtained by first reducing the ODE to first-order ODEs (see
Laboratory 4).
Let v = y′. Then v′ = y′′ = −ω20y = −4y. Also v(0) = y′(0) = 0. The following MATLAB program
implements the problem (with ω0 = 2).
function LAB05ex1
m = 1; % mass [kg]
k = 4; % spring constant [N/m]
omega0 = sqrt(k/m);
y0 = 0.1; v0 = 0; % initial conditions
[t,Y] = ode45(@f,[0,10],[y0,v0],[],omega0); % solve for 0<t<10
y = Y(:,1); v = Y(:,2); % retrieve y, v from Y
figure(1); plot(t,y,’b+-’,t,v,’ro-’); % time series for y and v
grid on;
%-----------------------------------------
function dYdt = f(t,Y,omega0)
y = Y(1); v = Y(2);
dYdt = [ v ; -omega0^2*y ];
Note that the parameter ω0 was passed as an argument to ode45 rather than set to its value ω0 = 2
directly in the funct ...

Chapter 14 Statics

To determine if an object is in equilibrium, there can be no resultant force or torque. This means the sum of all forces and sum of all torques must equal zero. To analyze equilibrium problems, one draws a free body diagram, chooses an axis of rotation where information is lacking, sums the torques and forces about that axis, setting each sum equal to zero to solve for unknowns. Common examples analyze beams, ropes, and physical systems with multiple supports to find tensions and reactive forces.

Chapter 14 Statics

To determine if an object is in equilibrium, there can be no resultant force or torque. This means the sum of all forces and sum of all torques must equal zero. To analyze equilibrium problems, one draws a free body diagram, chooses an axis of rotation where information is lacking, sums the torques and forces about that axis, setting each sum equal to zero to solve for unknowns. Common examples involve finding tensions in ropes or forces from supports of beams or structures.

Analysis of indeterminate beam by slopeand deflection method

Slope-deflection method ,Slope-deflection equations, equilibrium equation of
method, application to beams with and without joint translation and rotation, Sinking or yielding of support,

vector algebra - exercise on basics

This document discusses testing basic concepts in vector algebra. It provides examples of classifying quantities as scalars and vectors, identifying different vector properties like coinitial, equal and collinear vectors, and determining whether statements about vector properties are true or false. The goal is to solve problems involving the basics of vector algebra and assess confidence in doing so.

Struc lec. no. 1

The document summarizes key concepts in the theory of structures including:
- Types of loads, reactions, and supports
- Statically determinate beams, frames, arches, and trusses
- Relationship between loads, shear forces, and bending moments
- Concepts of stability, determinacy, and methods of analysis for solving equilibrium and conditional equations
Examples are provided to demonstrate solving for reactions, internal forces, and conditional equations for various statically determinate structures. Factors affecting stability and determinacy are also discussed.

Lecture 7 - Spring-Mass Problem.pptx

1) The document discusses the spring-mass problem and vibration of springs based on Hooke's law. It provides the equations of motion for an undamped and damped spring-mass system with and without an imposed force.
2) Examples are presented of an oscillating mass on an undamped and damped spring to illustrate the equations of motion.
3) Harmonic motion and resonance are introduced, with equations provided for undamped vibrations using Laplace transforms. The document discusses the cases when the natural and damping frequencies are equal or not equal.

Based on differences in electronegativity- how would you characterize.docx

Based on differences in electronegativity, how would you characterize the bonding in sulfur dioxide, SO2?
Essay answers are limited to about 500 words (3800 characters maximum, including spaces).
3800 Character(s) remaining
Solution
The atom with less electronegative is central atom and the atoms with more electronegtive are terminal atoms.
So, Sulphur with less electronegative is central atom and Oxygens are terminal atoms.
S ( Z = 16 ) 1st excited state electronic configuration is,
1s2 s2 2p6 3s2 3p3 3d1
S in its first exicted state undergoes sp2 hybirdisation due ot which it has 1 paired sp2 hybrid orbital and 2 unpaired sp2 hybrid orbitals. And also S has one p and one d orbital with unpaired electrons.
O (Z = 8) E.C is 1s2 2s2 2p4
Each O has two unpaired electrons in its valence shell.
The each sp2 unpaired hybrid orbital of S overlaps with one of the unpaired p - orbital ofeach oxygen along the internuclear axis to form single S - O sigma bond with each Oxygen.
The reamining unpaired pure p-orbitals of S overlaps with unpaired p - orbital of each oxygen perpendicular to the internuclear axis to form one pi bonds with each oxygen.
Still S ia remained with one lone pair of electons.
Due to this SO 2 is in bent shape and has the bond angle < 120 0
.

Balance the redox reaction by inserting the appropriate coefficients-.docx

Balance the redox reaction by inserting the appropriate coefficients. redox reaction: | Fe3+ + NO + H2O- Fe2+ + H+ + No;
Solution
Fe in Fe+3 has oxidation state of +3
Fe in Fe+2 has oxidation state of +2
So, Fe in Fe+3 is reduced to Fe+2
N in NO2- has oxidation state of +3
N in NO3- has oxidation state of +5
So, N in NO2- is oxidised to NO3-
Reduction half cell:
Fe+3 + 1e- --> Fe+2
Oxidation half cell:
NO2- --> NO3- + 2e-
Balance number of electrons to be same in both half reactions
Reduction half cell:
2 Fe+3 + 2e- --> 2 Fe+2
Oxidation half cell:
NO2- --> NO3- + 2e-
Lets combine both the reactions.
2 Fe+3 + NO2- --> 2 Fe+2 + NO3-
Balance Oxygen by adding water
2 Fe+3 + NO2- + H2O --> 2 Fe+2 + NO3-
Balance Hydrogen by adding H+
2 Fe+3 + NO2- + H2O --> 2 Fe+2 + NO3- + 2 H+
This is balanced chemical equation in acidic medium
Answer:
2 Fe 3+ + NO 2 - + H 2 O â€”> 2 Fe 2+ + 2 H + + NO 3 -
.

Auditing and Accounting Cases Investigating Issues of Fraud and Profes (2).docx

The auditor of BLMIS, David Friehling, failed to uphold the fundamental principles governing an audit by not exercising due care or maintaining professional skepticism. He disregarded his responsibility to independently verify BLMIS's assets, review revenue sources, examine bank accounts, verify liabilities, or confirm the purchase and custody of securities. Additionally, Friehling and his relatives had personal accounts with BLMIS, compromising his independence. In summary, Friehling showed a lack of intent to properly audit BLMIS and gather sufficient evidence, violating several key audit principles.

b- Impact testing was used to assess which property of the materials-.docx

b. Impact testing was used to assess which property of the materials? For metals, what crystal structure was more susceptible to the ductile-to-brittle transition temperature? Why was this the case?
Solution
Answer:-
The Imapct testing was used to to asses the required energy to fracture the material and it can be measure the toughness of the material. we can also known the yield strength of material by using impact testing but we dont have any analytical formual for. we can also calcualte the strain rate for its effect on fracture of material.
There is BCC and FCC type of crystal which is more susceptible to the ductile to brittle transiotion temperature.
the low carbon steel BCC metals, which become brittle at low temperature or at very high strain rate.
incase of FCC metals its remain ductiule at low temperature.
In metals there is plastic deformation at room temperature occurs due to dislocation motion. this dislocation motion required a stress to move and its depend on the atomic bonding which is a crustal structure.
In this case brittle fracture occurs at low temperature and ductile fracture occurs at high temperature. ductile-brittle transformation occured in between low and high temperature.
.

B- lonic Bonds- Some atoms have such a strong attraction for electrons.docx

Some atoms have a strong attraction for electrons, called electronegativity, that allows them to pull electrons away from other atoms to form ions. Atoms that gain electrons become negatively charged anions, while atoms that lose electrons become positively charged cations. Cations and anions are attracted to each other due to their opposite charges and form ionic compounds. Ionic bonds result from the electrostatic attraction between oppositely charged ions.

b- Does Rutherford-'s model predict the correct spectrum- Answer- Acco.docx

b. Does Rutherford\'s model predict the correct spectrum? Answer: According to Bohr, why did an atom not collapse in on itself while its electrons traveled around the nucleus? (1) 3. Answer: (1) 4. According to Bohr, what is happening in the atom when a photon of light is emitted? Answer: (1) 5. According to Bohr, what is happening in the atom when a photon of light is absorbed? Answer: 6. When you learned about the energy levels of hydrogen, an energy level diagram was introduced. On this diagram, the no energy level was represented. Go to the Hydrogen Atom simulation (Unit D), and complete the n 1 tono transition. Observe the energy state data. a. What is the energy of the n oo energy level? Answer b. If an electron is initially in the ground state, how much energy must be absorbed by the atom for its transition to the oo energy level? Answer:
Solution
3) In Bhor\'s model electrons are moving around the neucleous. Neucleus attract the electrons but still they do not fall on neucleus or they do not collapse because due to movement of electron a centrifugal force is also acting on the electron. This centrifugal force and attractive force of neucleus acts as centripetal force and thus electrons do not collapse.
4) According to Bhor when a photon of light is emitting from an atom the electrons are coming to their higher energy state to lower energy state.
5) Similarly when photon of light absorbed by the atom electrons of lower energy states jumps to higher energy state.
.

Auditors should perform audit procedures to identify and assess subseq.docx

Auditors should perform audit procedures to identify and assess subsequent events:
A.
Through year end
B.
Through issuance of the audit report
C.
Through the last day of field work
D.
For a reasonable period after year end
Solution
Option
.

Auditing and Accounting Cases Investigating Issues of Fraud and Profes.docx

Auditing and Accounting Cases Investigating Issues of Fraud and Professional Ethics 4th Edition
Case 1.4 Page 20
2. Provide one specific example of how Sunbeam violated the revenue recogni-tion principle in this situation. In your description, please identify a journal entry that may have been used by Sunbeam to commit the fraudulent act.
Solution
Answer: Revenue Recognition Principle: Tells accountants when to record revenue (only after it has been earned) and the amount of revenue to record (the cash value of what has been received).
Earnings process is complete IF service has been performed, goods have been delivered, or title has passed to the buyer
Persuasive evidence of arrangement for customer payment (Customer paid cash OR made a promise to pay)
The price is either fixed OR determinable (means: no uncertainties about the amount)
Collection is reasonably assured
In the case of Sunbeam:
.

AuditorSolution1) The auditors opinion on internal control is satisfie.docx

Auditor
Solution
1)
The auditorâ€™s opinion on internal control is satisfied, because the management has followed all the guidelines and has taken necessary steps for an effective internal control.
2)
The auditor has expressed an unqualified opinion on Hewlett Packardâ€™s financial statements, because the auditor observed during the audit that the income statement, cash flow statement, balance sheet and other financial statements show true and fair view figures. Unqualified opinion means the auditor believes that there is no fraud has committed by the client.
3)
The paragraph states that when there are permanent limitations like human fallibility, collusion etc. the auditor cannot give assurance whether the internal controls show the misstatements. Without having proper idea, if management or anyone estimates or forecasts the performance or quality of work to be done may lead to risk and the controls which are expected to follow in those estimates or forecasts may become inadequate or useless due change of conditions. Because we cannot expect same performance in every day and the policies and procedures are slowly become worst or useless.
.

Auditing and Accounting Cases Investigating Issues of Fraud and Profes (5).docx

Auditing and Accounting Cases Investigating Issues of Fraud and Professional Ethics 4th Edition
Case 1.3 Page 15
3. Consult Paragraph 67 of PCAOB Auditing Standard No. 12. Do you believe that Qwest had established an effective system of internal control over finan-cial reporting related to the presentation and disclosure of its nonrecurring revenue? Why or why not?
Solution
The revenue recognition principle states that revenues are recognized when realized or realizable and when earned; also, revenue from a transaction must meet both criteria to be recognized for recording in the financial statements.Â Â Revenues are realized when products are exchanged for cash or claims to cash, and are realizable when related assets received are readily convertible to cash.Â Â Revenues are earned when products are delivered or services are performed.Â Â These principles are important in defining revenue recognition, which helps auditors identify fraud and misstatements.Â Â Users of financial statements will have a better understanding of an entity
.

Athens- Inc-- has 9-000 shares of common stock outstanding and the fol.docx

Athens, Inc., has 9,000 shares of common stock outstanding and the following data has been collected:
Common stock, $10 par value, 20,000 shares authorized and 10,000 issued..........$100,000
Contributed capital in excess of pay value, common stock.... $50,000
Retained earnings........$25,000
Treasury stock...........411,500
The cost per share of the treasury stock is:
A $1.15
B $11.50
C $1.28
D $10.50
Solution
Ans B $ 11.50
.

At time 110 four processes (P1 P4) are waiting for a timeout signal-.docx

At time 110 four processes ( P1 â€“ P4 ) are waiting for a timeout signal.
They are scheduled to wake up respectively at times: 130, 145, 175, 210.
P2 wakes up and immediately issues a call to SetLTimer (tn, 40 ).
Show the new priority queue after the call. (call takes no time).
Process queue
Time Difference
Timer
Solution
Solution:
Timer = 30
.

At January 1- 2012- Lance Link Corporation had 100-000 shares of $1 pa.docx

At January 1, 2012, Lance Link Corporation had 100,000 shares of $1 par common stock and 1,000 shares of $100 par, 6%, cumulative preferred stock outstanding.
Express did not pay dividends in 2012 or 2013. In December 2014, the board of directors declared a $45,000 cash dividend to be paid on January 31.
Required:
Compute the total dividend to be paid to the preferred shareholders.
Compute the total dividend to be paid to the common shareholders
Show the December 31 journal entry to record the declaration of dividends.
Solution
Answer 1:
Answer 2:
Answer 3:
Profit and Loss A/c Dr. 45000
To Proposed Dividend 45000
(Being dividend declared)
.

At December 31- 2013- Hansen Corporation had 42-000 shares of common s.docx

At December 31, 2013, Hansen Corporation had 42,000 shares of common stock and 6,000 shares of 9%, $100 par cumulative preferred stock outstanding. No dividends were declared or paid in 2013. Net income was reported as $220,000. What is basic EPS? (Round your answer to 2 decimal places.)
Solution
Basic EPS is the amount of company\'s profit taht is available to its common stock shareholders. As we pay preferred shareholders before common stock holders , so we deduct the preferred dividend.
Basic EPS = (Net income - preferred dividend)/ weighted average number of shares
Net Income = $220000
Preferred dividend = 600000 x .09 = $54000
weighted number of shares ( assuming that all the shares were held by company for the whole year) = 42000
therefore,
Basic EPS = (220000 - 54000)/42000 = $3.95
.

Ch01 2

To solve differential equations of the form y' = ay - b:
1) Use methods of calculus to find the general solution, which is y = Ce^(at) + b/a, where C is a constant.
2) If initial conditions are given, set them equal to the general solution to determine the value of C, yielding a unique solution.
3) The equilibrium solution, where y' = 0, is found by setting y = b/a.

From the Front LinesOur robotic equipment and its maintenanc.docx

From the Front Lines
Our robotic equipment and its maintenance represent a fixed cost of $23,320 per month. The cost-effectiveness of robotic-assisted surgery is related to patient volume: With only 10 cases, the fixed cost per case is $2,332, and with 40 cases, the fixed cost per case is $583.
Source: Alemozaffar, Chang, Kacker, Sun, DeWolf, & Wagner (2013).
MATLAB sessions: Laboratory 5
MAT 275 Laboratory 5
The Mass-Spring System
In this laboratory we will examine harmonic oscillation. We will model the motion of a mass-spring
system with differential equations.
Our objectives are as follows:
1. Determine the effect of parameters on the solutions of differential equations.
2. Determine the behavior of the mass-spring system from the graph of the solution.
3. Determine the effect of the parameters on the behavior of the mass-spring.
The primary MATLAB command used is the ode45 function.
Mass-Spring System without Damping
The motion of a mass suspended to a vertical spring can be described as follows. When the spring is
not loaded it has length ℓ0 (situation (a)). When a mass m is attached to its lower end it has length ℓ
(situation (b)). From the first principle of mechanics we then obtain
mg︸︷︷︸
downward weight force
+ −k(ℓ − ℓ0)︸ ︷︷ ︸
upward tension force
= 0. (L5.1)
The term g measures the gravitational acceleration (g ≃ 9.8m/s2 ≃ 32ft/s2). The quantity k is a spring
constant measuring its stiffness. We now pull downwards on the mass by an amount y and let the mass
go (situation (c)). We expect the mass to oscillate around the position y = 0. The second principle of
mechanics yields
mg︸︷︷︸
weight
+ −k(ℓ + y − ℓ0)︸ ︷︷ ︸
upward tension force
= m
d2(ℓ + y)
dt2︸ ︷︷ ︸
acceleration of mass
, i.e., m
d2y
dt2
+ ky = 0 (L5.2)
using (L5.1). This ODE is second-order.
(a) (b) (c) (d)
y
ℓ
ℓ0
m
k
γ
Equation (L5.2) is rewritten
d2y
dt2
+ ω20y = 0 (L5.3)
c⃝2011 Stefania Tracogna, SoMSS, ASU
MATLAB sessions: Laboratory 5
where ω20 = k/m. Equation (L5.3) models simple harmonic motion. A numerical solution with ini-
tial conditions y(0) = 0.1 meter and y′(0) = 0 (i.e., the mass is initially stretched downward 10cms
and released, see setting (c) in figure) is obtained by first reducing the ODE to first-order ODEs (see
Laboratory 4).
Let v = y′. Then v′ = y′′ = −ω20y = −4y. Also v(0) = y′(0) = 0. The following MATLAB program
implements the problem (with ω0 = 2).
function LAB05ex1
m = 1; % mass [kg]
k = 4; % spring constant [N/m]
omega0 = sqrt(k/m);
y0 = 0.1; v0 = 0; % initial conditions
[t,Y] = ode45(@f,[0,10],[y0,v0],[],omega0); % solve for 0<t<10
y = Y(:,1); v = Y(:,2); % retrieve y, v from Y
figure(1); plot(t,y,’b+-’,t,v,’ro-’); % time series for y and v
grid on;
%-----------------------------------------
function dYdt = f(t,Y,omega0)
y = Y(1); v = Y(2);
dYdt = [ v ; -omega0^2*y ];
Note that the parameter ω0 was passed as an argument to ode45 rather than set to its value ω0 = 2
directly in the funct ...

Chapter 14 Statics

To determine if an object is in equilibrium, there can be no resultant force or torque. This means the sum of all forces and sum of all torques must equal zero. To analyze equilibrium problems, one draws a free body diagram, chooses an axis of rotation where information is lacking, sums the torques and forces about that axis, setting each sum equal to zero to solve for unknowns. Common examples analyze beams, ropes, and physical systems with multiple supports to find tensions and reactive forces.

Chapter 14 Statics

To determine if an object is in equilibrium, there can be no resultant force or torque. This means the sum of all forces and sum of all torques must equal zero. To analyze equilibrium problems, one draws a free body diagram, chooses an axis of rotation where information is lacking, sums the torques and forces about that axis, setting each sum equal to zero to solve for unknowns. Common examples involve finding tensions in ropes or forces from supports of beams or structures.

Analysis of indeterminate beam by slopeand deflection method

Slope-deflection method ,Slope-deflection equations, equilibrium equation of
method, application to beams with and without joint translation and rotation, Sinking or yielding of support,

vector algebra - exercise on basics

This document discusses testing basic concepts in vector algebra. It provides examples of classifying quantities as scalars and vectors, identifying different vector properties like coinitial, equal and collinear vectors, and determining whether statements about vector properties are true or false. The goal is to solve problems involving the basics of vector algebra and assess confidence in doing so.

Struc lec. no. 1

The document summarizes key concepts in the theory of structures including:
- Types of loads, reactions, and supports
- Statically determinate beams, frames, arches, and trusses
- Relationship between loads, shear forces, and bending moments
- Concepts of stability, determinacy, and methods of analysis for solving equilibrium and conditional equations
Examples are provided to demonstrate solving for reactions, internal forces, and conditional equations for various statically determinate structures. Factors affecting stability and determinacy are also discussed.

Lecture 7 - Spring-Mass Problem.pptx

1) The document discusses the spring-mass problem and vibration of springs based on Hooke's law. It provides the equations of motion for an undamped and damped spring-mass system with and without an imposed force.
2) Examples are presented of an oscillating mass on an undamped and damped spring to illustrate the equations of motion.
3) Harmonic motion and resonance are introduced, with equations provided for undamped vibrations using Laplace transforms. The document discusses the cases when the natural and damping frequencies are equal or not equal.

Ch01 2

Ch01 2

From the Front LinesOur robotic equipment and its maintenanc.docx

From the Front LinesOur robotic equipment and its maintenanc.docx

Chapter 14 Statics

Chapter 14 Statics

Chapter 14 Statics

Chapter 14 Statics

Analysis of indeterminate beam by slopeand deflection method

Analysis of indeterminate beam by slopeand deflection method

vector algebra - exercise on basics

vector algebra - exercise on basics

Struc lec. no. 1

Struc lec. no. 1

Lecture 7 - Spring-Mass Problem.pptx

Lecture 7 - Spring-Mass Problem.pptx

Based on differences in electronegativity- how would you characterize.docx

Based on differences in electronegativity, how would you characterize the bonding in sulfur dioxide, SO2?
Essay answers are limited to about 500 words (3800 characters maximum, including spaces).
3800 Character(s) remaining
Solution
The atom with less electronegative is central atom and the atoms with more electronegtive are terminal atoms.
So, Sulphur with less electronegative is central atom and Oxygens are terminal atoms.
S ( Z = 16 ) 1st excited state electronic configuration is,
1s2 s2 2p6 3s2 3p3 3d1
S in its first exicted state undergoes sp2 hybirdisation due ot which it has 1 paired sp2 hybrid orbital and 2 unpaired sp2 hybrid orbitals. And also S has one p and one d orbital with unpaired electrons.
O (Z = 8) E.C is 1s2 2s2 2p4
Each O has two unpaired electrons in its valence shell.
The each sp2 unpaired hybrid orbital of S overlaps with one of the unpaired p - orbital ofeach oxygen along the internuclear axis to form single S - O sigma bond with each Oxygen.
The reamining unpaired pure p-orbitals of S overlaps with unpaired p - orbital of each oxygen perpendicular to the internuclear axis to form one pi bonds with each oxygen.
Still S ia remained with one lone pair of electons.
Due to this SO 2 is in bent shape and has the bond angle < 120 0
.

Balance the redox reaction by inserting the appropriate coefficients-.docx

Balance the redox reaction by inserting the appropriate coefficients. redox reaction: | Fe3+ + NO + H2O- Fe2+ + H+ + No;
Solution
Fe in Fe+3 has oxidation state of +3
Fe in Fe+2 has oxidation state of +2
So, Fe in Fe+3 is reduced to Fe+2
N in NO2- has oxidation state of +3
N in NO3- has oxidation state of +5
So, N in NO2- is oxidised to NO3-
Reduction half cell:
Fe+3 + 1e- --> Fe+2
Oxidation half cell:
NO2- --> NO3- + 2e-
Balance number of electrons to be same in both half reactions
Reduction half cell:
2 Fe+3 + 2e- --> 2 Fe+2
Oxidation half cell:
NO2- --> NO3- + 2e-
Lets combine both the reactions.
2 Fe+3 + NO2- --> 2 Fe+2 + NO3-
Balance Oxygen by adding water
2 Fe+3 + NO2- + H2O --> 2 Fe+2 + NO3-
Balance Hydrogen by adding H+
2 Fe+3 + NO2- + H2O --> 2 Fe+2 + NO3- + 2 H+
This is balanced chemical equation in acidic medium
Answer:
2 Fe 3+ + NO 2 - + H 2 O â€”> 2 Fe 2+ + 2 H + + NO 3 -
.

Auditing and Accounting Cases Investigating Issues of Fraud and Profes (2).docx

The auditor of BLMIS, David Friehling, failed to uphold the fundamental principles governing an audit by not exercising due care or maintaining professional skepticism. He disregarded his responsibility to independently verify BLMIS's assets, review revenue sources, examine bank accounts, verify liabilities, or confirm the purchase and custody of securities. Additionally, Friehling and his relatives had personal accounts with BLMIS, compromising his independence. In summary, Friehling showed a lack of intent to properly audit BLMIS and gather sufficient evidence, violating several key audit principles.

b- Impact testing was used to assess which property of the materials-.docx

b. Impact testing was used to assess which property of the materials? For metals, what crystal structure was more susceptible to the ductile-to-brittle transition temperature? Why was this the case?
Solution
Answer:-
The Imapct testing was used to to asses the required energy to fracture the material and it can be measure the toughness of the material. we can also known the yield strength of material by using impact testing but we dont have any analytical formual for. we can also calcualte the strain rate for its effect on fracture of material.
There is BCC and FCC type of crystal which is more susceptible to the ductile to brittle transiotion temperature.
the low carbon steel BCC metals, which become brittle at low temperature or at very high strain rate.
incase of FCC metals its remain ductiule at low temperature.
In metals there is plastic deformation at room temperature occurs due to dislocation motion. this dislocation motion required a stress to move and its depend on the atomic bonding which is a crustal structure.
In this case brittle fracture occurs at low temperature and ductile fracture occurs at high temperature. ductile-brittle transformation occured in between low and high temperature.
.

B- lonic Bonds- Some atoms have such a strong attraction for electrons.docx

Some atoms have a strong attraction for electrons, called electronegativity, that allows them to pull electrons away from other atoms to form ions. Atoms that gain electrons become negatively charged anions, while atoms that lose electrons become positively charged cations. Cations and anions are attracted to each other due to their opposite charges and form ionic compounds. Ionic bonds result from the electrostatic attraction between oppositely charged ions.

b- Does Rutherford-'s model predict the correct spectrum- Answer- Acco.docx

b. Does Rutherford\'s model predict the correct spectrum? Answer: According to Bohr, why did an atom not collapse in on itself while its electrons traveled around the nucleus? (1) 3. Answer: (1) 4. According to Bohr, what is happening in the atom when a photon of light is emitted? Answer: (1) 5. According to Bohr, what is happening in the atom when a photon of light is absorbed? Answer: 6. When you learned about the energy levels of hydrogen, an energy level diagram was introduced. On this diagram, the no energy level was represented. Go to the Hydrogen Atom simulation (Unit D), and complete the n 1 tono transition. Observe the energy state data. a. What is the energy of the n oo energy level? Answer b. If an electron is initially in the ground state, how much energy must be absorbed by the atom for its transition to the oo energy level? Answer:
Solution
3) In Bhor\'s model electrons are moving around the neucleous. Neucleus attract the electrons but still they do not fall on neucleus or they do not collapse because due to movement of electron a centrifugal force is also acting on the electron. This centrifugal force and attractive force of neucleus acts as centripetal force and thus electrons do not collapse.
4) According to Bhor when a photon of light is emitting from an atom the electrons are coming to their higher energy state to lower energy state.
5) Similarly when photon of light absorbed by the atom electrons of lower energy states jumps to higher energy state.
.

Auditors should perform audit procedures to identify and assess subseq.docx

Auditors should perform audit procedures to identify and assess subsequent events:
A.
Through year end
B.
Through issuance of the audit report
C.
Through the last day of field work
D.
For a reasonable period after year end
Solution
Option
.

Auditing and Accounting Cases Investigating Issues of Fraud and Profes.docx

Auditing and Accounting Cases Investigating Issues of Fraud and Professional Ethics 4th Edition
Case 1.4 Page 20
2. Provide one specific example of how Sunbeam violated the revenue recogni-tion principle in this situation. In your description, please identify a journal entry that may have been used by Sunbeam to commit the fraudulent act.
Solution
Answer: Revenue Recognition Principle: Tells accountants when to record revenue (only after it has been earned) and the amount of revenue to record (the cash value of what has been received).
Earnings process is complete IF service has been performed, goods have been delivered, or title has passed to the buyer
Persuasive evidence of arrangement for customer payment (Customer paid cash OR made a promise to pay)
The price is either fixed OR determinable (means: no uncertainties about the amount)
Collection is reasonably assured
In the case of Sunbeam:
.

AuditorSolution1) The auditors opinion on internal control is satisfie.docx

Auditor
Solution
1)
The auditorâ€™s opinion on internal control is satisfied, because the management has followed all the guidelines and has taken necessary steps for an effective internal control.
2)
The auditor has expressed an unqualified opinion on Hewlett Packardâ€™s financial statements, because the auditor observed during the audit that the income statement, cash flow statement, balance sheet and other financial statements show true and fair view figures. Unqualified opinion means the auditor believes that there is no fraud has committed by the client.
3)
The paragraph states that when there are permanent limitations like human fallibility, collusion etc. the auditor cannot give assurance whether the internal controls show the misstatements. Without having proper idea, if management or anyone estimates or forecasts the performance or quality of work to be done may lead to risk and the controls which are expected to follow in those estimates or forecasts may become inadequate or useless due change of conditions. Because we cannot expect same performance in every day and the policies and procedures are slowly become worst or useless.
.

Auditing and Accounting Cases Investigating Issues of Fraud and Profes (5).docx

Auditing and Accounting Cases Investigating Issues of Fraud and Professional Ethics 4th Edition
Case 1.3 Page 15
3. Consult Paragraph 67 of PCAOB Auditing Standard No. 12. Do you believe that Qwest had established an effective system of internal control over finan-cial reporting related to the presentation and disclosure of its nonrecurring revenue? Why or why not?
Solution
The revenue recognition principle states that revenues are recognized when realized or realizable and when earned; also, revenue from a transaction must meet both criteria to be recognized for recording in the financial statements.Â Â Revenues are realized when products are exchanged for cash or claims to cash, and are realizable when related assets received are readily convertible to cash.Â Â Revenues are earned when products are delivered or services are performed.Â Â These principles are important in defining revenue recognition, which helps auditors identify fraud and misstatements.Â Â Users of financial statements will have a better understanding of an entity
.

Athens- Inc-- has 9-000 shares of common stock outstanding and the fol.docx

Athens, Inc., has 9,000 shares of common stock outstanding and the following data has been collected:
Common stock, $10 par value, 20,000 shares authorized and 10,000 issued..........$100,000
Contributed capital in excess of pay value, common stock.... $50,000
Retained earnings........$25,000
Treasury stock...........411,500
The cost per share of the treasury stock is:
A $1.15
B $11.50
C $1.28
D $10.50
Solution
Ans B $ 11.50
.

At time 110 four processes (P1 P4) are waiting for a timeout signal-.docx

At time 110 four processes ( P1 â€“ P4 ) are waiting for a timeout signal.
They are scheduled to wake up respectively at times: 130, 145, 175, 210.
P2 wakes up and immediately issues a call to SetLTimer (tn, 40 ).
Show the new priority queue after the call. (call takes no time).
Process queue
Time Difference
Timer
Solution
Solution:
Timer = 30
.

At January 1- 2012- Lance Link Corporation had 100-000 shares of $1 pa.docx

At January 1, 2012, Lance Link Corporation had 100,000 shares of $1 par common stock and 1,000 shares of $100 par, 6%, cumulative preferred stock outstanding.
Express did not pay dividends in 2012 or 2013. In December 2014, the board of directors declared a $45,000 cash dividend to be paid on January 31.
Required:
Compute the total dividend to be paid to the preferred shareholders.
Compute the total dividend to be paid to the common shareholders
Show the December 31 journal entry to record the declaration of dividends.
Solution
Answer 1:
Answer 2:
Answer 3:
Profit and Loss A/c Dr. 45000
To Proposed Dividend 45000
(Being dividend declared)
.

At December 31- 2013- Hansen Corporation had 42-000 shares of common s.docx

At December 31, 2013, Hansen Corporation had 42,000 shares of common stock and 6,000 shares of 9%, $100 par cumulative preferred stock outstanding. No dividends were declared or paid in 2013. Net income was reported as $220,000. What is basic EPS? (Round your answer to 2 decimal places.)
Solution
Basic EPS is the amount of company\'s profit taht is available to its common stock shareholders. As we pay preferred shareholders before common stock holders , so we deduct the preferred dividend.
Basic EPS = (Net income - preferred dividend)/ weighted average number of shares
Net Income = $220000
Preferred dividend = 600000 x .09 = $54000
weighted number of shares ( assuming that all the shares were held by company for the whole year) = 42000
therefore,
Basic EPS = (220000 - 54000)/42000 = $3.95
.

File- LineScrabbleDriver-java -U 4- in-Desktop-Guess this is the death.docx

File: LineScrabbleDriver.java /U 4/ in/Desktop/Guess this is the death of me-GRASP CSD uava. 87% )- 12:19 Q Project Settings Tools Window Help import java.util.Scanner: import java.io.* public class LineScrabbleDriver public static void main(String argsl]) try string fileName; Scanner scan new Scanner (System.in); System.out.println(\"Enter a file name please fileName = scan. nextLine(); Linescrabble anew Linescrabble(fileName) a.readLines) a.doLine(); e.consequence) catch (IOException e) System.out.println( \"Please enter a file name whose file has strings.\" Echo.java LineScrabbleDriver.java LineScrabble.java GRASP Messages | Runl/O | Interactions LineScrabbleDriver . java: 14: doLine( java . 1ang.String) intineserabble cannotbe applied to a dolineOi l error Compile Messages Stop Clear Copy Line:15 Col:1 Code:97 Top:1
Solution
In your LineScrabble.java
instead of public void doLine(String c)
replace with public void processLine(String c)
then it will execute successfully
.

Fill out the following table for four different IT systems- Note two.docx

Fill out the following table for four different IT systems.
Â·Â Â Â Â Â Â Â Note two enterprise systems they connect with and their connection type.
Â·Â Â Â Â Â Â Â Note two security vulnerabilities the connection may have and two to four ways each vulnerability could be potentially exploited.
Additional Comments:
An example row has been entered into the table in blue. This is only an example and should not limit what you do.
Keep in mind that enterprise systems cover a certain task in the enterprise (HR, CRM, Identity Management, etc.). They are not the components of a system (such as servers).
Connections can often be a direct connection/pipe, a file, a common database, or something else.
The vulnerability is what would make the connection vulnerable to an attack.
The related risk is an attack that could target the weakness.
Solution
When securing the modern enterprise, consider that IT systems do not operate alone. Securing them involves securing their interfaces with other systems as well. It is important to know the different interconnections each system may have.two security vulnerabilities the system may have and 2 to 4 ways each vulnerability could be potentially exploited. IT System Target System Connection Type Possible Security.
Vulnerability Related Risk
1. WMOS
2. WCS
3. network(database)
4.Incorrect accessibility
5.System Failures
6. Users accessing items not allowed
7.Users not having access to needed items
8.Loss of data and reboot time
9.Downtime of production
A first step to developing an enterprise security plan is to identify the specific vulnerabilities and related risks facing an organization. This list should be fairly exhaustive. Many vulnerability and threat pairs will not make the final cut for remediation, but an organization can only properly prioritize these if it has fully covered all of the risks.
.

Figure 9-3 shows an energy diagram for the dissolving of an ionic comp.docx

Figure 9.3 shows an energy diagram for the dissolving of an ionic compound. Discuss what factors are going to tend to make this process more favorable-in other words would likely make the com pound more soluble.
Solution
Factors that increases the solubility of ionic compounds:
Higher the solvent-solute interaction, higher the solubility of ionic compounds in the solvent. As polar solvent dissolve polar solute, ionic compounds are more soluble in polar solvent like water. Polar solvent strongly attract the ions of the solid compound.
Common ion effect decrease the solubility of the ionic compounds if the salts ion is already present in the solvent. For example CaSO?(s) ? CaÂ²?(aq) + SO?Â²?(aq)
If either of Ca 2+ and SO 4 2- is already present in the solvent the solubility of CaSO4 in the solvent decrease due to common ion effect.
For an endothermic reaction increase in temperature increases the solubility of the ionic compound in the solvent as predicted by Le Chatelier Principle.
For an exothermic reaction increase in temperature decreases the solubility of ionic compound in the according to Le Chatelier Principle.
.

FakeArrayList first - new FakeArrayl_ist(3)- first-set(2- 7-1)- first-.docx

FakeArrayList first = new FakeArrayl_ist(3); first.set(2, 7.1); first.set(l, 4.9); FakeArrayList second = new FakeArrayList(5); second.set(0, 9.1); second.set(1, 8.4);
Solution
class main_stack_frame
{ int identifier;
int address;
float contents;
public:
void set(int id,float cont)
{identifier=id;
contents=cont;
address=address;
}
}
class heap extends main_stack_frame
{ int identifier1;
int address1;
float contents1;
public:
void set(int id1,float cont1)
{identifier1=id1;
contents1=cont1;
address1=address1;
}
}
class test
{ public static void main(String args[])
main_stack_frame first= new main_stack_frame(3); // object of stack_frame class
heap second = new heap(5);// object of heap class
first.set(2,7.1);// calling set function of main_stack_frame
first.set(1,4.9)
second.set(0,9.1);// invoking set methd of heap class
second.set(1,8.4);
}
}
.

Explain why workplace spirituality seems to be an important concern-So.docx

Explain why workplace spirituality seems to be an important concern.
Solution
A culture that is most likely to shape high ethical standards is high in risk tolerance, low to moderate in aggressiveness, and focuses on means as well as outcomes. An innovative culture is characterized by the following: challenge and involvement, freedom, trust and openness, idea time, playfulness/humor, conflict resolution, debates, and risk-taking. A customer-responsive culture has six characteristics: employees who are outgoing and friendly; few rigid rules, procedures, and regulations; widespread use of empowerment; clear roles and expectations; and employees who are conscientious in their desire to please the customer.
Workplace spirituality is important for the following reasons: employees are looking for ways to counterbalance the stresses and pressures of a turbulent pace of life; people are looking for involvement and connection that they often don
.

Explanation of the similarities and differences between the Income Sta.docx

Explanation of the similarities and differences between the Income Statement and
Operating Activities section of the Statement of Cash Flows
a. What type of information is each statement conveying? Be specific and provide at least
two examples.
Solution
Cash flow from operating activities is a section of the cash flow statement that provides information regarding the cash-generating abilities of a company\'s core activities.
Example:
Cash flow from operating activities is generally calculated according to the following formula:
Cash Flow from Operating Activities = Net income + Noncash Expenses + Changes in Working Capital
The noncash expenses are usually the depreciation and/or amortization expenses listed on the firm\'sincome statement.
A statement of cash flows typically breaks out a company\'s cash sources and uses for the period into three categories: cash flows from operations, cash flows from investing activities, and cash flows from financing activities. Cash flows from operations primarily measures the cash-generating abilities of the company\'s core operations rather than from its ability to raise capital or purchase assets
The income statement is one of the three primary financial statements used to assess a company
.

Based on differences in electronegativity- how would you characterize.docx

Based on differences in electronegativity- how would you characterize.docx

Balance the redox reaction by inserting the appropriate coefficients-.docx

Balance the redox reaction by inserting the appropriate coefficients-.docx

Auditing and Accounting Cases Investigating Issues of Fraud and Profes (2).docx

Auditing and Accounting Cases Investigating Issues of Fraud and Profes (2).docx

b- Impact testing was used to assess which property of the materials-.docx

b- Impact testing was used to assess which property of the materials-.docx

B- lonic Bonds- Some atoms have such a strong attraction for electrons.docx

B- lonic Bonds- Some atoms have such a strong attraction for electrons.docx

b- Does Rutherford-'s model predict the correct spectrum- Answer- Acco.docx

b- Does Rutherford-'s model predict the correct spectrum- Answer- Acco.docx

Auditors should perform audit procedures to identify and assess subseq.docx

Auditors should perform audit procedures to identify and assess subseq.docx

Auditing and Accounting Cases Investigating Issues of Fraud and Profes.docx

Auditing and Accounting Cases Investigating Issues of Fraud and Profes.docx

AuditorSolution1) The auditors opinion on internal control is satisfie.docx

AuditorSolution1) The auditors opinion on internal control is satisfie.docx

Auditing and Accounting Cases Investigating Issues of Fraud and Profes (5).docx

Auditing and Accounting Cases Investigating Issues of Fraud and Profes (5).docx

Athens- Inc-- has 9-000 shares of common stock outstanding and the fol.docx

Athens- Inc-- has 9-000 shares of common stock outstanding and the fol.docx

At time 110 four processes (P1 P4) are waiting for a timeout signal-.docx

At time 110 four processes (P1 P4) are waiting for a timeout signal-.docx

At January 1- 2012- Lance Link Corporation had 100-000 shares of $1 pa.docx

At January 1- 2012- Lance Link Corporation had 100-000 shares of $1 pa.docx

At December 31- 2013- Hansen Corporation had 42-000 shares of common s.docx

At December 31- 2013- Hansen Corporation had 42-000 shares of common s.docx

File- LineScrabbleDriver-java -U 4- in-Desktop-Guess this is the death.docx

File- LineScrabbleDriver-java -U 4- in-Desktop-Guess this is the death.docx

Fill out the following table for four different IT systems- Note two.docx

Fill out the following table for four different IT systems- Note two.docx

Figure 9-3 shows an energy diagram for the dissolving of an ionic comp.docx

Figure 9-3 shows an energy diagram for the dissolving of an ionic comp.docx

FakeArrayList first - new FakeArrayl_ist(3)- first-set(2- 7-1)- first-.docx

FakeArrayList first - new FakeArrayl_ist(3)- first-set(2- 7-1)- first-.docx

Explain why workplace spirituality seems to be an important concern-So.docx

Explain why workplace spirituality seems to be an important concern-So.docx

Explanation of the similarities and differences between the Income Sta.docx

Explanation of the similarities and differences between the Income Sta.docx

220711130083 SUBHASHREE RAKSHIT Internet resources for social science

Internet resources for social science

How to Fix [Errno 98] address already in use

This slide will represent the cause of the error “[Errno 98] address already in use” and the troubleshooting steps to resolve this error in Odoo.

Information and Communication Technology in Education

(𝐓𝐋𝐄 𝟏𝟎𝟎) (𝐋𝐞𝐬𝐬𝐨𝐧 2)-𝐏𝐫𝐞𝐥𝐢𝐦𝐬
𝐄𝐱𝐩𝐥𝐚𝐢𝐧 𝐭𝐡𝐞 𝐈𝐂𝐓 𝐢𝐧 𝐞𝐝𝐮𝐜𝐚𝐭𝐢𝐨𝐧:
Students will be able to explain the role and impact of Information and Communication Technology (ICT) in education. They will understand how ICT tools, such as computers, the internet, and educational software, enhance learning and teaching processes. By exploring various ICT applications, students will recognize how these technologies facilitate access to information, improve communication, support collaboration, and enable personalized learning experiences.
𝐃𝐢𝐬𝐜𝐮𝐬𝐬 𝐭𝐡𝐞 𝐫𝐞𝐥𝐢𝐚𝐛𝐥𝐞 𝐬𝐨𝐮𝐫𝐜𝐞𝐬 𝐨𝐧 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐫𝐧𝐞𝐭:
-Students will be able to discuss what constitutes reliable sources on the internet. They will learn to identify key characteristics of trustworthy information, such as credibility, accuracy, and authority. By examining different types of online sources, students will develop skills to evaluate the reliability of websites and content, ensuring they can distinguish between reputable information and misinformation.

INTRODUCTION TO HOSPITALS & AND ITS ORGANIZATION

The document discuss about the hospitals and it's organization .

BPSC-105 important questions for june term end exam

BPSC-105 important questions for june term end exam

220711130088 Sumi Basak Virtual University EPC 3.pptx

Virtual University

BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 8 - CẢ NĂM - FRIENDS PLUS - NĂM HỌC 2023-2024 (B...

BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 8 - CẢ NĂM - FRIENDS PLUS - NĂM HỌC 2023-2024 (B...Nguyen Thanh Tu Collection

https://app.box.com/s/nrwz52lilmrw6m5kqeqn83q6vbdp8yzpskeleton System.pdf (skeleton system wow)

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إضغ بين إيديكم من أقوى الملازم التي صممتها
ملزمة تشريح الجهاز الهيكلي (نظري 3)
💀💀💀💀💀💀💀💀💀💀
تتميز هذهِ الملزمة بعِدة مُميزات :
1- مُترجمة ترجمة تُناسب جميع المستويات
2- تحتوي على 78 رسم توضيحي لكل كلمة موجودة بالملزمة (لكل كلمة !!!!)
#فهم_ماكو_درخ
3- دقة الكتابة والصور عالية جداً جداً جداً
4- هُنالك بعض المعلومات تم توضيحها بشكل تفصيلي جداً (تُعتبر لدى الطالب أو الطالبة بإنها معلومات مُبهمة ومع ذلك تم توضيح هذهِ المعلومات المُبهمة بشكل تفصيلي جداً
5- الملزمة تشرح نفسها ب نفسها بس تكلك تعال اقراني
6- تحتوي الملزمة في اول سلايد على خارطة تتضمن جميع تفرُعات معلومات الجهاز الهيكلي المذكورة في هذهِ الملزمة
واخيراً هذهِ الملزمة حلالٌ عليكم وإتمنى منكم إن تدعولي بالخير والصحة والعافية فقط
كل التوفيق زملائي وزميلاتي ، زميلكم محمد الذهبي 💊💊
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220711130082 Srabanti Bag Internet Resources For Natural Science

Internet resources for natural science

Oliver Asks for More by Charles Dickens (9)

Oliver-Asks-for-More by Charles Dickens

220711130100 udita Chakraborty Aims and objectives of national policy on inf...

Aims and objectives of national policy on information and communication technology(ICT) in school education in india

Pharmaceutics Pharmaceuticals best of brub

First year pharmacy
Best for u

KHUSWANT SINGH.pptx ALL YOU NEED TO KNOW ABOUT KHUSHWANT SINGH

INDIA`S OWN LITERARY GENIUS MR.KHUSHWANT SINGH WAS TRULY A VERY BRAVE SOUL AND WAS AWARDED WITH THE MAGIC OF WORDS BY GOD.

Accounting for Restricted Grants When and How To Record Properly

In this webinar, member learned how to stay in compliance with generally accepted accounting principles (GAAP) for restricted grants.

Elevate Your Nonprofit's Online Presence_ A Guide to Effective SEO Strategies...

Whether you're new to SEO or looking to refine your existing strategies, this webinar will provide you with actionable insights and practical tips to elevate your nonprofit's online presence.

Geography as a Discipline Chapter 1 __ Class 11 Geography NCERT _ Class Notes...

Geography as discipline

Andreas Schleicher presents PISA 2022 Volume III - Creative Thinking - 18 Jun...

Andreas Schleicher, Director of Education and Skills at the OECD presents at the launch of PISA 2022 Volume III - Creative Minds, Creative Schools on 18 June 2024.

Standardized tool for Intelligence test.

ASSESSMENT OF INTELLIGENCE USING WITH STANDARDIZED TOOL

220711130083 SUBHASHREE RAKSHIT Internet resources for social science

220711130083 SUBHASHREE RAKSHIT Internet resources for social science

How to Fix [Errno 98] address already in use

How to Fix [Errno 98] address already in use

Information and Communication Technology in Education

Information and Communication Technology in Education

INTRODUCTION TO HOSPITALS & AND ITS ORGANIZATION

INTRODUCTION TO HOSPITALS & AND ITS ORGANIZATION

BPSC-105 important questions for june term end exam

BPSC-105 important questions for june term end exam

220711130088 Sumi Basak Virtual University EPC 3.pptx

220711130088 Sumi Basak Virtual University EPC 3.pptx

BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 8 - CẢ NĂM - FRIENDS PLUS - NĂM HỌC 2023-2024 (B...

BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 8 - CẢ NĂM - FRIENDS PLUS - NĂM HỌC 2023-2024 (B...

skeleton System.pdf (skeleton system wow)

skeleton System.pdf (skeleton system wow)

220711130082 Srabanti Bag Internet Resources For Natural Science

220711130082 Srabanti Bag Internet Resources For Natural Science

Oliver Asks for More by Charles Dickens (9)

Oliver Asks for More by Charles Dickens (9)

220711130100 udita Chakraborty Aims and objectives of national policy on inf...

220711130100 udita Chakraborty Aims and objectives of national policy on inf...

Pharmaceutics Pharmaceuticals best of brub

Pharmaceutics Pharmaceuticals best of brub

KHUSWANT SINGH.pptx ALL YOU NEED TO KNOW ABOUT KHUSHWANT SINGH

KHUSWANT SINGH.pptx ALL YOU NEED TO KNOW ABOUT KHUSHWANT SINGH

Accounting for Restricted Grants When and How To Record Properly

Accounting for Restricted Grants When and How To Record Properly

Elevate Your Nonprofit's Online Presence_ A Guide to Effective SEO Strategies...

Elevate Your Nonprofit's Online Presence_ A Guide to Effective SEO Strategies...

RESULTS OF THE EVALUATION QUESTIONNAIRE.pptx

RESULTS OF THE EVALUATION QUESTIONNAIRE.pptx

Geography as a Discipline Chapter 1 __ Class 11 Geography NCERT _ Class Notes...

Geography as a Discipline Chapter 1 __ Class 11 Geography NCERT _ Class Notes...

The basics of sentences session 7pptx.pptx

The basics of sentences session 7pptx.pptx

Andreas Schleicher presents PISA 2022 Volume III - Creative Thinking - 18 Jun...

Andreas Schleicher presents PISA 2022 Volume III - Creative Thinking - 18 Jun...

Standardized tool for Intelligence test.

Standardized tool for Intelligence test.

- 1. Fart A stant If you ignore the weight of the bar itself, how far from the left end of the barbell is the center of gravity? Express your answer to three significant figures and include the appropriate units A 1.80-m-long barbell has a 20.0 kg weight on its left end and a 36.0 kg weight on its right end You may want to review (Pages 204- 205) HA Submit Previous Answers Request Answer X Incorrect; Try Again Part B Where is the center of gravity if the 7.00 kg mass of the barbell itself is taken into account? Express your answer to three significant figures and include the appropriate units HA dValue Units Solution Part A : If we ignore the weight of the bar itself, how far from the left end of a barbell is the center of gravity? we know that, x cg = (m 1 x 1 + m 2 x 2 ) / (m 1 + m 2 ) x cg = [(20 kg) (0 m) + (35 kg) (1.8 m)] / [(20 kg) + (35 kg)] x cg = (63 kg.m) / (55 kg) x cg = 1.14 m Part B : Where is the center of gravity if the 7 kg mass of a barbell itself is taken into account? we know that, x cg = (m 1 x 1 + m 2 x 2 + m 3 x 3 ) / (m 1 + m 2 + m 3 ) x cg = [(20 kg) (0 m) + (35 kg) (1.8 m) + (7 kg) (1.8 m / 2)] / [(20 kg) + (35 kg) + (7 kg)] x cg = [(63 kg.m) + (6.3 kg.m)] / (62 kg) x cg = [(69.3 kg.m) / (62 kg)] x cg = 1.11 m