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2 years ago

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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.)

1 answer:

Igoryamba2 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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Leviafan [203]

Answer:

32 C > 32 F > 32 K

Explanation:

32 F, 32 C, 32 K

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3 years ago

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siniylev [52]

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The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

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The distance from the ball to the wall = 4 m

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The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

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Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

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2 years ago

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Answer:

The block's mass should be $3m$

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Now new mass $M$ is added to the system,

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$A = v \sqrt{\frac{m +M }{k} }$

Here, given in question frequency is reduced to half so we can write,

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Where $f =$ frequency of system before mass is added, $f' =$ frequency of system after mass is added.

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