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Svet_ta [14]
3 years ago
14

A tennis ball traveling horizontally at a speed of 40 m/s hits a wall and rebounds in the opposite direction. The time Interval

for the collision is about 0.013 s, and the mass of the ball Is 0.059 kg. Assume that the ball rebounds at the same speed.
(a) Determine the initial and final momenta of the ball. Assume the initial velocity direction is positive.

(b) Determine the change In momentum of the ball.

(c) Determine the average force exerted by the wall on the ball.
Physics
1 answer:
liubo4ka [24]3 years ago
3 0

Answer

given,

initial velocity of the ball, u = 40 m/s

final velocity of the ball, v= -40 m/s

time of contact = 0.013 s

mass of the ball = 0.059 Kg

a) initial momentum

   P₁ = m u = 0.059 x 40 = 2.36 kg.m/s

  final momentum  

   P₂ = m v = 0.059 x (-40) = -2.36 kg.m/s

b) change in momentum

      Δ P = P₂- P₁

      Δ P = -2.36 - 2.36

     Δ P = -4.72 kg.m/s

c) Average force

 average force exerted by the ball is equal to change in momentum per unit time.

     F =\dfrac{\Delta P}{t}

     F =\dfrac{-4.72}{0.013}

            F = -363 N

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A driver traveling at 30.0 m/s presses the brakes decelerates at -8.80 m/s^2. How much time, in second, does it take her to come
Kruka [31]

Answer:

It takes her 3.409 seconds to make a full stop.

Explanation:

The time it takes to make a full stop can be determined by the equation of velocity for a Uniformly Accelerated Rectilinear Motion:

v_{f} = v_{i} + at  (1)

Where v_{f} is the final velocity, v_{i} is the initial velocity, a is the acceleration and t is the time.

Equation (1) can be rewritten in terms of t:

v_{f} - v_{i} = at    

t = \frac{v_{f} - v_{i}}{a}  (2)  

For this particular case the final velocity will be equal to zero (v_{f} = 0):

t = \frac{0 m/s - 30.0 m/s}{-8.80 m/s^{2}}

t = 3.409 s

So it takes her 3.409 seconds to make a full stop.

4 0
3 years ago
Can anybody help me solve this problem? Thank you so much!
ser-zykov [4K]
Please ignore my comment -- mass is not needed, here is how to solve it. pls do the math

at bottom box has only kinetic energy
ke = (1/2)mv^2
v = initial velocity
moving up until rest work done = Fs
F = kinetic fiction force = uN = umg x cos(a)
s = distance travel = h/sin(a)
h = height at top
a = slope angle
u = kinetic fiction
work = Fs = umgh x cot(a)
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3 years ago
Peter’s body supplies a force of 500 N to run up a 10-m hill in 10 s. How much power is involved in Peter’s run up the hill? Exp
shutvik [7]

Answer: 500 Watts

Explanation:

Power P is the speed with which work W is done. Its unit is Watts (W), being 1 W=\frac{1 Joule}{1 s}.

Power is mathematically expressed as:

P=\frac{W}{t} (1)

Where t is the time during which work W  is performed.

On the other hand, the Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  It is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter (1J=(1N)(1m)=Nm  ).

When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (2)

In this case, we have the following data:

F=500 N

d=10 m

t=10 s

So, let's calculate the work done by Peter and then find how much power is involved:

From (2):

W=(500 N)(10 m) (3)

W=5000 J (4)

Substituting (4) in (1):

P=\frac{5000 J}{10 s} (5)

Finally:

P=500 W

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