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Klio2033 [76]
3 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]3 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= (105 kg * 9.8 m.s^2 - 5*105) / 105 kg

a = 9.3 m/s

Hope this helps
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Explanation:

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Question I. SUHU ULT
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Feathers are great thermal insulators.

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Feathers are great thermal insulators. The loose structure of down feathers traps air.

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Human beings discovered that down feathers are good for insulation long ago. For example, documents from the 1600s show that Russian merchants sold “bird down" to the Dutch hundreds of years ago.

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3 0
3 years ago
In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e
Zanzabum

Answer:

The  velocity is  v_t  =  0.02175 \  m/s

Explanation:

From the question we are told that

   The  mass of the bullet is  m_b  =  0.024 \  kg

    The initial speed of the bullet is  u_b  =  1200 \  m/s

   The mass of the target is  m_t  =  320 \  kg

    The  initial velocity of target is  u_t  =  0  \ m/s

    The  final velocity of the bullet is  is  v_b  =  910 \  m/s

   

Generally according to the law of momentum conservation we have that

      m_b *  u_b  +  m_t *  u_t  =  m_b *  v_b  +  m_t  *  v_t

=>   0.024  *  1200  +  320 *  0  =  0.024 *  910   +  320  *  v_t

=>    v_t  =  0.02175 \  m/s

3 0
3 years ago
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