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Usimov [2.4K]
3 years ago
11

A hockey puck with a mass of 0.17 kg is traveling to the right along the ice at 15 m/s. it strikes a second hockey puck with a m

ass 0.11 kg. the first hockey puck comes to rest after the collision. What is the velocity of the second hockey puck after the collision? ( Round your answer to the nearest integer.)
Physics
1 answer:
Igoryamba3 years ago
3 0

Answer:

v_2 = 23.182m/s

Explanation:

Given

m_1 = 0.17kg

m_2 = 0.11kg

u_1 = 15m/s -- Initial Velocity of the first

u_2 = 0m/s -- Initial Velocity of the second

v_1 = 0m/s -- Final Velocity of the first

Required

Determine the final velocity of the second hockey puck (v_2)

This question illustrates law of conservation of momentum and it'll be solved using the following formula:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 ---- law of conservation of momentum

Substitute in the right values

0.17 * 15 + 0.11*0 = 0.17 * 0 + 0.11*v_2

0.17 * 15 + 0 = 0 + 0.11*v_2

2.55 = 0.11*v_2

Solve for v2

v_2 = 2.55/0.11

v_2 = 23.182m/s

<em>Hence, the final velocity of the second hockey puck is 23.182m/s</em>

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A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?
Pie

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

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A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1 2 and t = 1 s? Use G
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There is one mistake in the question.The Correct question is here

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

Answer:

y(1s) - y(1/2s) =  - 3.675 m  

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Explanation:

Given data

time=1/2 sec to 1 sec

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To find

Distance

Solution

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So

acceleration(-g)=  dv/dt

Solve it

dv  =  a dt

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v=  dy/dt

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dy =  ( v₀ - gt ) dt

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y1s - y1/2s  = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

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