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
1 answer:
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,
So, the magnitude of the impulse that the player applies to the ball is 9.34 kg-m/s.
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The spring constant is 4 N/m
Explanation:
When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:
where
F is the magnitude of the spring applied
k is the spring constant
x is the elongation of the spring, relative to its equilibrium position
For the spring in this problem, we have:
F = 0.12 N (force applied)
x = 3 cm = 0.03 m (elongation of the spring)
Therefore, we can solve the formula for k to find the spring constant:
Learn more about forces:
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