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3 years ago

3

When the lever is pulled, 2 kg of carbon dioxide is ejected at a speed of 60 m/s. The remaining mass of the person, chair, and c

ylinder is 80 kg.
After the ejection, how fast will the chair be moving?

Select the best answer from the choices provided.

a)2.0 m/s

b)0.5 m/s

c)1.0 m/s

d)1.5 m/s

1 answer:

vova2212 [387]3 years ago

5 0

Answer:

<em>d)1.5 m/s</em>

Explanation:

<em>Momentum:</em> Momentum can be defined as the product of mass and velocity.

The S.I unit of momentum is kgm/s. Momentum is a vector quantity.

Momentum of the carbon dioxide = momentum of the chair.

mv = MU............................. Equation 1

making U the subject of the equation above,

U = mv/M.............................. Equation 2

where m = mass of the carbon-dioxide, v = velocity of the carbon-dioxide, M = mass of the chair, U = velocity of the chair

<em>Given: m = 2 kg, v = 60 m/s, M = 80 kg</em>

Substituting these values into equation 2

U = 2×60/80

U = 120/80

U = 1.5 m/s

<em>Therefore the speed of the chair after ejection = 1.5 m/s</em>

<em>The right option is d)1.5 m/s</em>

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