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andrezito [222]

2 years ago

10

A model air rocket with a mass of 50.g is free to travel along a horizontal track. It begins from rest. After 2.0s, the rocket h

as lost 10% of its mass by expelling air at an average velocity of 53.m⋅s−1. What is the velocity of the rocket at that moment

1 answer:

katen-ka-za [31]2 years ago

7 0

Answer:

$v_r=5.89\ m.s^{-1}$

Explanation:

Given:

mass of rocket, $m_r=50\ g$

time of observation, $t=2\ s$

mass lost by the rocket by expulsion of air, $m_a=10\%\ of m_r=5\ g$

velocity of air, $v_a=53\ m.s^{-1}$

<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)

$m_a.v_a=(m_r-m_a)\times v_r$

$5\times 53=(50-5)\times v_r$

$v_r=5.89\ m.s^{-1}$

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Answer:

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2 years ago

Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with the message 0.44 kg is on the table. Becc

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Julius competes in the hammer throw event. The hammer has a mass of 7.26 kg and is 1.215 m long. What is the centripetal force o

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In the circular motion of the hammer, the centripetal force is given by
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2 years ago

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A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

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Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

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$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

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We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

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which is approximately 41º west of north.

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2 years ago

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<span>
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2 years ago