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zloy xaker [14]
3 years ago
5

When a 58g tennis ball is served, it accelerates from rest to a constant speed of 36 m/s. The impact with the racket gives the b

all a constant acceleration over a distance of 35 cm. What is the magnitude of the net force acting on the ball?
Physics
1 answer:
inysia [295]3 years ago
5 0
We first calculate the acceleration on the ball using:
2as = v² - u²; u = 0 because ball is initially at rest
a = (36)²/(2 x  0.35)
a = 1850 m/s²
F = ma
F = 0.058 x 1850
= 107.3 Newtons
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A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin
pychu [463]

a) See graph in attachment

b) The suvat equation to use is v_f^2 - v_i^2 = 2as

c) The acceleration is a=\frac{v_f^2-v_i^2}{2s}

d) The acceleration is 1.25 m/s^2

e) The time needed is 8 s

Explanation:

a)

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is x_i = 0, the origin

- The  final position of the boat is x_f = 200 m

- The initial velocity of the boat is v_i = 20.0 m/s

- The final velocity of the boat is v_f = 30.0 m/s

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

b)

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

v_i = 20.0 m/s, the initial velocity

v_f = 30.0 m/s, the final velocity

s = x_f - x_i = 200 m, the  displacement of the boat

Therefore, the equation that best can be use to find the acceleration is

v_f^2 - v_i^2 = 2as

where

a is the acceleration

c)

Now we have to solve the equation

v_f^2 - v_i^2 = 2as

In order to find the acceleration.

This can be done by dividing both terms by 2s: this way, we find

\frac{v_f^2-v_i^2}{2s}=\frac{2as}{2s}

And so the acceleration is

a=\frac{v_f^2-v_i^2}{2s}

d)

Now we can use the equation found in part c) in order to find the acceleration.

We have the following data:

v_i = 20.0 m/s, the initial velocity

v_f = 30.0 m/s, the final velocity

s = x_f - x_i = 200 m, the  displacement of the boat

And substituting into the equation,

a=\frac{30^2-20^2}{2(200)}=1.25 m/s^2

e)

In order to find the time it takes the boat to travel the given distance, we can use the following suvat equation:

v_f = v_i + at

where:

v_i is the initial velocity

v_f is the final velocity

a is the acceleration

t is the time

Here we have:

v_i = 20.0 m/s

v_f = 30.0 m/s

a=1.25 m/s^2

Solving for t, we find:

t=\frac{v_f-v_i}{a}=\frac{30-20}{1.25}=8 s

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