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Anni [7]
3 years ago
8

The change in momentum of an object is equal to the ____________ that acts on it.

Physics
2 answers:
meriva3 years ago
3 0

Answer : The change in momentum of an object is equal to the impulse that acts on it.

Explanation :

Change in momentum : The change in momentum of an object is the product of the mass and the change in velocity of an object.

The formula of change in momentum is,

\Delta p=m\times \Delta v

Impulse : An impulse of an object is the product of the force applied on an object and the change in time. Impulse is also equivalent to the change in momentum of  an object.

J=F\times \Delta t

Proof :

J=F\times \Delta t\\\\J=(m\times a)\times \Delta t\\\\J=m\times (a\times \Delta t)\\\\J=m\times \Delta v=\Delta p

Hence, the change in momentum of an object is equal to the impulse that acts on it.

ipn [44]3 years ago
3 0
It's equal to the FORCE that acts on it
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How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer:
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According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:

W=\Delta K

Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:

K=\frac{1}{2}mv^2

Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:

\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\  \\ \therefore K\approx532,346J \end{gathered}

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2 years ago
Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a
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a) 32 kg m/s

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p_i = m u = (8 kg)(2 m/s)=16 kg m/s

The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

p_f = p_{fB}+p_{fS}

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p_i = p_f\\p_i = p_{fB}+p_{fS}\\p_{fS}=p_i -p_{fB}=16 kg m/s -(-16 kg m/s)=32 kg m/s


b) -32 kg m/s

The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s


c) 64 N

The change in momentum is equal to the product between the average force and the time of the interaction:

\Delta p=F \Delta t

Since we know \Delta t=0.5 s, we can find the magnitude of the force:

F=\frac{\Delta p}{\Delta t}=\frac{-32 kg m/s}{0.5 s}=-64 N

The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.


d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

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