|
QUIZ 6 |
|||||||||||||||
|
Exercises
and Problems from Conceptual Physics: Note that answers to odd numbered
problems are listed in the paperback supplement "Practicing
Physics." |
|||||||||||||||
|
Ch
8 |
exercises |
1 |
4 |
6 |
30 |
33 |
41 |
47 |
|
|
|
|
|
|
|
|
|
problems |
|
2 |
5 |
6 |
10 |
|
|
|
|
|
|
|
|
|
|
Ch
9 |
exercises |
|
4 |
11 |
12 |
14 |
19 |
26 |
28 |
37 |
47 |
51 |
|
||
|
|
problems |
1 |
2 |
3 |
5 |
|
|
|
|
|
|
|
|
||
|
Hints:
HINTS WILL BE POSTED SOON….. CHECK BACK LATER!!! |
|||||||||||||||
|
|
|||||||||||||||
|
exercises
|
|||||||||||||||
| 4. Sue's . See fig. 8.2 | |||||||||||||||
| 6. varying linear speed. See fig. 8.2 | |||||||||||||||
| 30. The weight of the beam acts at the geometric center of the beam. The torque of the boy = the torque due to the beam weight : (lever arm of boy) x (weight of boy) = (lever arm of beam weight) x (beam weight.).. The boy places the pivot at a location so that the above equation is true. Note that the lever arm of boy = distance between the pivot and the boy ....Note also that the lever arm of beam = distance between the center of the beam and the pivot. | |||||||||||||||
| 33. At the center of the earth. This is like the center of mass of a ring being at the geometric center of the ring. | |||||||||||||||
|
problems |
|||||||||||||||
|
2.
speed = radius x angular speed. angular speed = 6.28/30 s, where 6.28 =
2 x pi, where pi = 3.14 |
|||||||||||||||
| 6. (a) torque = lever arm x force (b) set (0.25 m) x ( 80 N) = (0.10 m) x ( new force) . Find the new force. | |||||||||||||||
|
10.
See page 151. Use
angular momentum = mvr.
Find the ratio of (mvr for the earth around the sun) to (mvr for the
moon around the earth.) ie. find ( mvr for the earth around the sun)/(mvr for
the moon around the earth.)...Use the Physical Data Table at the back of the
book. v = radius x angular speed, where angular speed = 6.28/(time for the
orbit) Background ![]() Actual Picture ![]() |
|||||||||||||||
|
|
|||||||||||||||
|
exercises |
|||||||||||||||
|
4. force is the same if they have the same mass: force = mg. |
|||||||||||||||
| 11. Nearer the moon because the moon has the smaller mass. | |||||||||||||||
| 14. The same from Newton's Third Law in Chapter 5. | |||||||||||||||
| 19. Your weight would go down if the earth expanded. Your weight would up of the earth shrunk. See the formula in problem 3 below. d = distance from the surface to the center of the earth. If d increases, then mg decreases. | |||||||||||||||
| 28. There is a force of gravity on the pencil due to the earth. But you are falling with the pencil, so the pencil hovers. | |||||||||||||||
|
problems |
|||||||||||||||
|
3.
mg = GMm/d2 . Cancel out the m. |
|||||||||||||||
| 5. mg = GMm/d2 , where d = 2x 105 m + earth radius . Find g. | |||||||||||||||
|
|
|||||||||||||||
|
|
|||||||||||||||