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3 years ago

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An astronaut whose mass is 80 kg carries an empty oxygen tank with a mass of 10 kg. The astronaut throws the tank away with a sp

eed of 2.0 m/s. With what velocity does the astronaut start to move off into space?

1 answer:

yan [13]3 years ago

6 0

Answer:

The speed of the astronaut is 0.25 m/s in the opposite direction to the oxygen tank.

Explanation:

Using the conservation of linear momentum, we have:

$p_{initial}=p_{final}$

The initial momentum is zero because both of them are at rest, so we just have a final momentum.

$0=m_{astro}V_{f-astro}+m_{tank}V_{f-tank}$

$m_{astro}V_{f-astro}=-m_{tank}V_{f-tank}$

$V_{f-astro}=-\frac{V_{f-tank}m_{tank}}{m_{astro}}$

$V_{f-astro}=-\frac{2*10}{80}$

$V_{f-astro}=-0.25\: m/s$

Therefore, the speed of the astronaut is 0.25 m/s in the opposite direction to the oxygen tank.

I hope it helps you!

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

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substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

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The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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