Answer:
<u>Displacement (km)</u>
Explanation:
The y axis is the vertical axis pointing up and down. This is labeled as the the displacement (km) in the graph.
The answer is to this question is B
Answer:
F=248.5W N
Explanation:
Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

Since we start counting at 0s and the ball departs from rest, this is just 
So we can write:

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

Substituting our values:

Where the average force exerted has been written it terms of the tennis ball's weight W.
Answer:
a) 0.147 N
b) 9.408 N
c) 9.261 N
Explanation:
The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force
. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.
Answer:
T = 0.017s
Explanation:
period is the time it takes a particle to make one oscillation
An electric current is periodic in nature
The current reaches 3.8A ten times.
So there must have been 10 cycles (10 periods) in 0.17s. let 'T' be the period:

t is the total time interval
n is the number of oscillations

10T = 0.17
T = 0.17/10 = 0.017s