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storchak [24]
3 years ago
6

A volleyball is spiked so that its incoming velocity of +2.09 m/s is changed to an outgoing velocity of -24.6 m/s. The mass of t

he volleyball is 0.350 kg. What is the magnitude of the impulse that the player applies to the ball?
Physics
1 answer:
Oksana_A [137]3 years ago
7 0

Answer:

Impulse, J = 9.34 kg-m/s

Explanation:

Given that,

Mass of the volleyball, m = 0.35 kg

Initial velocity, u = +2.09 m/s

Final speed, v = -24.6 m/s

We need to find the magnitude of the impulse that the player applies to the ball. We know that impulse is equal to the change in momentum . So,

J=m(v-u)\\\\J=0.35\times (-24.6-2.09)\\\\J=-9.34\ kg-m/s

So, the magnitude of the impulse that the player applies to the ball is 9.34 kg-m/s.

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A coffee filter of mass 1.5 grams dropped from a height of 3 m reaches the ground with a speed of 0.7 m/s. How much kinetic ener
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<h3 />

Given:

Mass of a coffee filter, m = 1.5 g

Height from which it is dropped, h = 3 m

Speed at ground, v = 0.7 m/s

Initially, the coffee filter has potential energy. It is given by :

P =mgh

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Finally, it will have kinetic energy. It is given by :

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E= \frac{1}{2}×1.4 × 10⁻³ × (0.7)²

E = 0.000343 J

The  kinetic energy Kair did the air molecules gain from the falling coffee filter is :

E = 0.000343 -  0.0441

  = 0.0437 J

So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J

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Can a machine like a lever, produce a greater force than what you put into it? Can it increase the energy that you put into it?
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