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kenny6666 [7]
3 years ago
13

A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the pers

on is 68.0 kg, the mass of the skateboard is 4.10 kg, and the mass of the brick is 2.50 kg. If the person throws the brick forward (in the direction they are facing) with a speed of 21.0 m/s relative to the skateboard and we ignore friction, determine the recoil speed of the person and the skateboard, relative to the ground.
Physics
1 answer:
jek_recluse [69]3 years ago
4 0

Answer:

0.74 m/s

Explanation:

From the question,

We apply the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Since the skateboard, the person and the brick where stationary, therefore, the total momentum before collision is 0

0 = Total momentum after collision

(m+M)V + m'v = 0

Where m = mass of the  skateboard, M = mass of the person, m' = mass of the brick, V = recoil velocity of the person and the skateboard, v =  velocity of the brick

make V the subject of the equation above

V = -m'v/(m+M)................... Equation 1

Given: m = 4.10 kg, M = 68.0 kg, m' = 2.50 kg, v = 21.0 m/s.

Substitute these values into equation 1

V = -(2.5×21)/(68+2.5)

V = 52.50/70.5

V = 0.74 m/s

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Answer:

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Where:

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b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

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A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

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I hope it helps you!                                                                  

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