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Klio2033 [76]
2 years ago
5

A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back t

oward your friend. Which of the following options will maximize your speed right after your interaction with the ball? You should catch the ball. You should deflect the ball back toward your friend at the same speed with which it hit your hand. You should let the ball go past you without touching it. It doesn’t matter—your speed is the same regardless of what you do. You should deflect the ball back toward your friend at half the speed with which it hit your hand.
Physics
1 answer:
baherus [9]2 years ago
6 0

Answer:

Explanation:

My speed after the interaction will depend upon the impulse the ball will make on me . Now impulse can be expressed as follows

Impulse = change in momentum

change in momentum in the ball will be maximum when the ball bounces back with the same velocity which can be shown as follows

change in momentum = mv - ( - mv ) = 2mv

So when ball is bounced back with same velocity , it suffers greatest impulse from my hand . In return ,  it reacts with the same impulse on my hand pushing me with greatest impulse according to third law of motion. this maximizes my speed after the interaction.

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A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
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\sum \tau = F*d

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A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
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Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

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the formula for the time period of the pendulum is given by

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where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

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By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

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Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

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