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Leto [7]
3 years ago
6

A bowling ball of mass 5 kg hits a wall going 7 m/s and rebounds at a speed of 2 m/s. What was the impulse applied to the bowlin

g ball
Physics
2 answers:
LekaFEV [45]3 years ago
6 0
The impulse applied to the bowling ball is 45 kg m/s
gizmo_the_mogwai [7]3 years ago
3 0

Explanation :

Given that,

Mass of the ball, m = 5 kg

Initial velocity of the ball, u = 7 m/s

Final speed of the ball, v = 2 m/s

We know that the impulse applied to an object is equal to the change in momentum.

J=\Delta p

J=p_f-p_i

J=m(v-u)

J=5\ kg(2\ m/s-7\ m/s)

J = -25 kgm/s

Hence, this is the required solution.

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<u>Hello and Good Morning/Afternoon</u>:

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<u>To balance this equation</u>:

⇒ must ensure that there is an equal number of elements on both sides of the equation at all times

<u>Let's start balancing:</u>

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              ⇒ but only so far one on the right side

         C<em>₂H₅OH + __O₂ →  2CO₂ + __ H₂O</em>

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               ⇒ but only so far two on the right side

         C<em>₂H₅OH + __O₂ →  2CO₂ + 3H₂O</em>

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<u>Let's check and make sure we got the answer:</u>

                           C<em>₂H₅OH + 3O₂ →  2CO₂ + 3H₂O</em>

<em>                 2 Carbon                ⇔                    2 Carbon</em>

<em>                 6 Hydrogen            ⇔                   6 Hydrogen</em>

<em>                 7 Oxygen                ⇔                   7 oxygen</em>

<u>Thefore the coefficients in order are</u>:

  ⇒ 1, 3, 2, 3

<u>Answer: 1,3,2,3</u>

Hope that helps!

#LearnwithBrainly<em>                      </em>

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