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Physics Assignment 2

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Assignment: * Exercises 3 and 29: 1. When a ball rolls down an inclined plane it gains speed because of gravity. When rolling up, it loses speed because of gravity. Why doesn't gravity play a role when it rolls on a horizontal surface?
While rolling level a ball does not roll with or against the vertical force of gravity, it neither speeds up nor slows down. The rolling ball maintains a constant speed; this happens because friction overcomes the ball’s inertia and brings it to a stop. The horizontal component of gravity is zero. 2. What is the acceleration of a car that moves at a steady velocity of 100km/hr for 100 seconds? Explain your answer, and state why this question is an exercise in careful reading as well as in Physics.
The acceleration is 0. There is no change of velocity in those 100 seconds time interval. The word you have to pay attention to is “steady” that is why is can exercise in careful reading and in Physics because you would try to do the math but at the end the answer would be wrong because the car never accelerated. * Problems 1,5,8 and 10 1. Find the net force produced by a 30-N force and a 20-N force in each of the following cases: a. Both forces act in the same direction.
30+20= 50N b. The two forces act in different directions.
30-20=10N
2. A vehicle changes its velocity from 100 km/h to a dead stop in 10 s. Show that the acceleration in stopping is -10 km/h x s.
VF-Vi= 0-100
VT/t = (0-100)/10s = -10 km/h x s 3. A ball is thrown straight up with enough speed so that it is in the air for several seconds. c. What is the velocity of the ball when it reaches its highest point?
Velocity is 0 at highest point. d. What is its velocity 1 s before it reaches its highest point?
9.8m/s
e. What is the change in its velocity during this 1-s interval?
-9.8m/s
f. What is its velocity 1 s after it reaches its highest point?
-9.8m/s
g. What is the change in velocity during this 1-s interval?
-9.8m/s
h. What is the change in velocity during the 2-s interval from 1 s before the highest point to 1 s before the highest point to 1 s after the highest point?
19.6m/s
i. What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity?
At any point acceleration is 9.8 m/s2 – Gravitational Constant 4. An electrically charged particle accelerates uniformly from rest to speed v while traveling a distance x. j. Show that the acceleration of the particle is a=v2/2x. d=initial velocity + final velocity/2 x time d=0+v/2*t x= vt/2 t=2x/v a= v/t a=v/(2x/v) a=v2/2x k. If the particle starts from rest and reaches a speed of 1.8x107 m/s over a distance of 0.10 m, show that its acceleration is 1.6 x1015 m/s2. a = v^2/2x = (1.8 x 107)2/ (0.2) = 1.6x1015 m/s2…...

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