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3 years ago

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A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the grou

nd exerts a +55 N force on the ball.

1 answer:

borishaifa [10]3 years ago

7 0

The final velocity of the ball is +9.25 m/s

Explanation:

We can solve the problem by using the impulse theorem, which states that the impulse exerted on the ball (product of the force exerted and the duration of the collision) is equal to the change in momentum of the ball:

$F\Delta t = m(v-u)$

where

F is the force exerted

$\Delta t$ is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem, we have

m = 0.060 kg

u = -32 m/s

$\Delta t = 45 ms = 0.045 s$

F = +55 N

Therefore we can solve for v to find the final velocity:

$v=u+\frac{F\Delta t}{m}=-32 + \frac{(55)(0.045)}{0.060}=+9.25 m/s$

Learn more about change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

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A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

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Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

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Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

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