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# Sci 1010 chapter 2

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### Sci 1010 chapter 2

1. 1. Chapter 2
2. 2.  Physics is concerned with the basic principles that describe how the universe works.  Physics deals with matter, motion, force, and energy. Intro
3. 3.  Classical mechanics  Waves and sounds  Thermodynamics  Electromagnetism  Quantum mechanics  Atomic and nuclear physics  Relativity Intro
4. 4.  Motion is everywhere – walking, driving, flying, etc.  This chapter focuses on definition/discussion of: speed, velocity, and acceleration.  There are two basic kinds of motion: ◦ Straight line ◦ Circular Intro
5. 5.  Position – the location of an object ◦ A reference point must be given in order to define the position of an object  Motion – an object is undergoing a continuous change in position  Description of Motion – the time rate of change of position ◦ A combination of length and time describes motion Section 2.1
6. 6.  In Physical Science ‘speed’ and ‘velocity’ have different (distinct) meanings.  Speed is a scalar quantity, with only magnitude ◦ A car going 80 km/h  Velocity is a vector, and has both magnitude & direction ◦ A car going 80 km/h north Section 2.2
7. 7.  Vector quantities may be represented by arrows. The length of the arrow is proportional to magnitude. 40 km/h 80 km/h -40 km/h -80 km/h Section 2.2 Vectors
8. 8. Note that vectors may be both positive and negative. Section 2.2
9. 9. ◦ ( where ∆ means ‘change in’) ◦ Over the entire time interval, speed is an average  Distance – the actual path length traveled  Instantaneous Speed – the speed of an object at an instant of time (∆t is very small) ◦ Glance at a speedometer Section 2.2 distance traveled time to travel distance • Average Speed = • v = d/t or v = ∆d/∆t
10. 10. Section 2.2
11. 11.  Velocity is similar to speed except a direction is involved.  Displacement – straight line distance between the initial and final position w/ direction toward the final position  Instantaneous velocity – similar to instantaneous speed except it has direction Section 2.2 displacement total travel time • Average velocity =
12. 12. Displacement is a vector quantity between two points. Distance is the actual path traveled. Section 2.2
13. 13.  Describe the speed of the car above.  GIVEN: d = 80 m & t = 4.0 s  SOLVE: d/t = 80 m/4.0 s = 20 m/s = average speed  Velocity would be described as 20 m/s in the direction of motion (east?) Section 2.2 • EQUATION: v = d/t
14. 14.  How far would the car above travel in 10s?  EQUATION: v = d/t  REARRANGE EQUATION: vt = d  SOLVE: (20 m/s) (10 s) = 200 m Section 2.2
15. 15.  GIVEN: ◦ Speed of light = 3.00 x 108 m/s = (v) ◦ Distance to earth = 1.50 x 108 km = (d)  EQUATION: ◦ v = d/t  rearrange to solve for t  t = d/v  t = 0.500 x 103 s = 5.00 x 102 s = 500 seconds Section 2.2 • SOLVE: t = d/v = 1.50 x 1011 m 3.00 x 108 m/s
16. 16. Section 2.2
17. 17.  GIVEN: ◦ t = 365 days (must convert to hours) ◦ Earth’s radius (r) = 9.30 x 107 miles  CONVERSION: ◦ t = 365 days x (24h/day) = 8760 h  MUST FIGURE DISTANCE (d):  d = ? Recall that a circle’s circumference = 2πr  d=2πr (and we know π and r!!)  Therefore  d = 2πr = 2 (3.14) (9.30 x 107 mi) Section 2.2
18. 18.  SOLVE EQUATION:  ADJUST DECIMAL POINT: Section 2.2 • d/t = 2 (3.14) (9.30 x 107 mi) = 0.00667 x 107 mi/h 8760 h • v = d/t • v = avg. velocity = 6.67 x 104 mi/h = 66,700 mi/h
19. 19.  Radius of the Earth = RE = 4000 mi  Time = 24 h  Diameter of Earth = 2πr = 2 (3.14) (4000mi) ◦ Diameter of Earth = 25,120 mi  v = d/t = (25,120 mi)/24h = 1047 mi/h Section 2.2
20. 20.  Changes in velocity occur in three ways: ◦ Increase in magnitude (speed up) ◦ Decrease in magnitude (slow down) ◦ Change direction of velocity vector (turn)  When any of these changes occur, the object is accelerating.  Faster the change  Greater the acceleration  Acceleration – the time rate of change of velocity Section 2.3
21. 21.  A measure of the change in velocity during a given time period Section 2.3 • Avg. acceleration = change in velocity time for change to occur • Units of acceleration = (m/s)/s = m/s2 • In this course we will limit ourselves to situations with constant acceleration. • a = = ∆v t vf – vo t (vf = final & vo = original)
22. 22.  As the velocity increases, the distance traveled by the falling object increases each second. Section 2.3
23. 23.  A race car starting from rest accelerates uniformly along a straight track, reaching a speed of 90 km/h in 7.0 s. Find acceleration.  GIVEN: vo = 0, vf = 90 km/h, t = 7.0s  WANTED: a in m/s2 Section 2.3 • vf = 90 km/h = 25 m/s)( / /278.0 hkm sm • a = = = 3.57 m/s2 vf –vo t 25 m/s – 0 7.0 s
24. 24.  Rearrange this equation:  at = vf – vo  vf = vo + at (solved for final velocity)  This equation is very useful in computing final velocity. Section 2.3 vf – vo t • Remember that a =
25. 25.  If the car in the preceding example continues to accelerate at the same rate for three more seconds, what will be the magnitude of its velocity in m/s at the end of this time (vf )?  a = 3.57 m/s2 (found in preceding example)  t = 10 s  vo = 0 (started from rest)  Use equation: vf = vo + at  vf = 0 + (3.57 m/s2 ) (10 s) = 35.7 m/s Section 2.3
26. 26. Section 2.3 Acceleration (+) Deceleration (-)
27. 27.  Special case associated with falling objects  Vector towards the center of the earth  Denoted by “g”  g = 9.80 m/s2 Section 2.3
28. 28. 0 10 20 30 40 50 60 70 0 2 4 6 8 Time in seconds Velocity The increase in velocity is linear. Section 2.3
29. 29.  If frictional effects (air resistance) are neglected, every freely falling object on earth accelerates at the same rate, regardless of mass. Galileo is credited with this idea/experiment.  Astronaut David Scott demonstrated the principle on the moon, simultaneously dropping a feather and a hammer. Each fell at the same acceleration, due to no atmosphere & no air resistance. Section 2.3
30. 30.  Galileo is also credited with using the Leaning Tower of Pisa as an experiment site. Section 2.3
31. 31.  d = ½ gt2  This equation will compute the distance (d) an object drops due to gravity (neglecting air resistance) in a given time (t). Section 2.3
32. 32.  A ball is dropped from a tall building. How far does the ball drop in 1.50 s?  GIVEN: g = 9.80 m/s2 , t = 1.5 s  SOLVE: d = ½ gt2 = ½ (9.80 m/s2 ) (1.5 s)2 = ½ (9.80 m/s2 ) (2.25 s2 ) = ?? m Section 2.3
33. 33.  A ball is dropped from a tall building. How far does the ball drop in 1.50 s?  GIVEN: g = 9.80 m/s2 , t = 1.5 s  SOLVE: d = ½ gt2 = ½ (9.80 m/s2 ) (1.5 s)2 = ½ (9.80 m/s2 ) (2.25 s2 ) = Section 2.3 11.0 m
34. 34.  What is the speed of the ball in the previous example 1.50 s after it is dropped?  GIVEN: g = a = 9.80 m/s2 , t = 1.5 s  SOLVE:  vf = vo + at = 0 + (9.80 m/s2 )(1.5 s)  Speed of ball after 1.5 seconds = 14.7 m/s Section 2.3
35. 35.  Distance is proportional to t2 – Remember that (d = ½ gt2 )  Velocity is proportional to t – Remember that (vf = at ) Section 2.3
36. 36. 0 50 100 150 200 0 2 4 6 8 Velocity Distance Time in seconds Section 2.3
37. 37.  Acceleration due to gravity occurs in BOTH directions. ◦ Going up (-) ◦ Coming down (+)  The ball returns to its starting point with the same speed it had initially. vo = vf Section 2.3
38. 38.  Although an object in uniform circular motion has a constant speed, it is constantly changing directions and therefore its velocity is constantly changing directions. ◦ Since there is a change in direction there is a change in acceleration.  What is the direction of this acceleration?  It is at right angles to the velocity, and generally points toward the center of the circle. Section 2.4
39. 39.  Supplied by friction of the tires of a car  The car remains in a circular path as long as there is enough centripetal acceleration. Section 2.4
40. 40.  This equation holds true for an object moving in a circle with radius (r) and constant speed (v).  From the equation we see that centripetal acceleration increases as the square of the speed.  We also can see that as the radius decreases, the centripetal acceleration increases. Section 2.4 v2 r • ac =
41. 41.  Determine the magnitude of the centripetal acceleration of a car going 12 m/s on a circular track with a radius of 50 m. Section 2.4
42. 42.  Determine the magnitude of the centripetal acceleration of a car going 12 m/s on a circular track with a radius of 50 m.  GIVEN: v = 12 m/s, r = 50 m  SOLVE: Section 2.4 • ac = = = 2.9 m/s2 v2 r (12 m/s)2 50m
43. 43.  Compute the centripetal acceleration in m/s2 of the Earth in its nearly circular orbit about the Sun.  GIVEN: r = 1.5 x 1011 m, v = 3.0 x 104 m/s  SOLVE: Section 2.4 • ac = = = v2 r (3.0 x 104 m/s)2 1.5 x 1011 m 9.0 x 108 m2 /s2 1.5 x 1011 m • ac = 6 x 10-3 m/s2
44. 44.  An object thrown horizontally combines both straight-line and vertical motion each of which act independently.  Neglecting air resistance, a horizontally projected object travels in a horizontal direction with a constant velocity while falling vertically due to gravity. Section 2.5
45. 45.  An object thrown horizontally will fall at the same rate as an object that is dropped. Multiflash photograph of two balls Section 2.5 Copyright © Standard HMCO copyright line
46. 46. Section 2.5
47. 47. Combined Horz/Vert. Components Vertical Component Horizontal Component += Section 2.5
48. 48. Section 2.5
49. 49. Section 2.5
50. 50. Section 2.5
51. 51.  Angle of release  Spin on the ball  Size/shape of ball/projectile  Speed of wind  Direction of wind  Weather conditions (humidity, rain, snow, etc)  Field altitude (how much air is present)  Initial horizontal velocity (in order to make it out of the park before gravity brings it down, a baseball must leave the bat at a high velocity) Section 2.5
52. 52.  d = ½at2 (distance traveled, starting from rest)  d = ½gt2 (distance traveled, dropped object)  g = 9.80 m/s2 = 32 ft/s2 (acceleration, gravity)  vf = vo + at (final velocity with constant a)  ac = v2 /r (centripetal acceleration) Review ∆v t vf –vo t • a = = (constant acceleration) • v = d/t (average speed)