A ball is dropped and bounces off the floor. Its speed is the same immediately before and immediately after the collision. Which
1 answer:
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Using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.
so then you sub what you have (v and lambada) in the formula.
then multiply the frequency(f) by the given wavelength and then solve for f
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.
so then you sub what you have (v and lambada) in the formula.
then multiply the frequency(f) by the given wavelength and then solve for f
Answer:
0.074m/s
Explanation:
We need the formula for conservation of momentum in a collision, this equation is given by,
Where,
= mass of ball
= mass of the person
= Velocity of ball before collision
= Velocity of the person before collision
= velocity of ball afer collision
= velocity of the person after collision
We know that after the collision, as the person as the ball have both the same velocity, then,
Re-arrenge to find ,
Our values are,
= 0.425kg
= 12m/s
= 68.5kg
= 0m/s
Substituting,
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<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>