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2 years ago

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A basketball of mass 0.23kg is thrown horizontally against a rigid vertical wall with a velocity of 20m/s. It rebounds with a ve

locity of 15m/s. Calculate the impulse of the force of the wall on the basketball.

1 answer:

Anna11 [10]2 years ago

7 0

Answer:

$8.1\:\mathrm{Ns}$

Explanation:

The impulse-momentum theorem gives the impulse on an object to be equal to the change in momentum of that object. Since mass is maintained, the change in momentum of the basketball is:

$\Delta p = m\Delta v$ , where $m$ is the mass of the basketball and $\Delta v$ is the change in velocity.

Since the basketball is changing direction, its total change in velocity is:

$\Delta v = 20-(-15)=35\:\mathrm{m/s}$ .

Therefore, the basketball's change in momentum is:

$\Delta p = m\Delta v = 0.23\cdot 35= 8.05=8.1\:\mathrm{kg\cdot m/s}$ .

Thus, the impulse on the basketball is $\fbox{$8.1\:\mathrm{Ns}$}$ (two significant figures).

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Answer:

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3 years ago

A hiker walks 20.51 m at 33.16 degrees. What is the Y component of his displacement?

Serhud [2]

Answer:

<em>The y component of his displacement is 11.22 meters</em>

Explanation:

<u>Components of the displacement</u>

The displacement is a vector because it has a magnitude and a direction. Let's suppose a displacement has a magnitude r and a direction θ, measured with respect to the positive x-direction. The horizontal component of the displacement is calculated by:

$x=r\cos\theta$

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$y=r\sin\theta$

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$y=20.51\sin(33.16^\circ)$

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3 years ago

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saveliy_v [14]

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3 years ago

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3 years ago