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2 years ago

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A person is pushing a box. The net external force on the 60-kg box is stated to be 90 N. If the force of friction opposing the m

otion is 30 N, what is the acceleration of the box?a. 0.66 m/s2b. 1.5 m/s2c. 2.0 m/s2

1 answer:

Tems11 [23]2 years ago

5 0

Answer:

$b.1.5m/s^2$

Explanation:

We are given that

Mass of box=60kg

Net external force applied on the box=90 N

Friction force =30N

We have to find the acceleration of the box.

We know that

Net external force=ma

Substitute the values then we get

$90=60a$

$a=\frac{90}{60}=1.5m/s^2$

Hence, option b is true.

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V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

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m*g*cos(θ) - 0 = m*vf^2 / R

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vf^2 = 1.45*32.2*cos(34)

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ΔK.E = ΔP.E

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( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))

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How many nanoseconds are there in 1.90 yr ?<br> Express your answer using three significant figures.

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Answer:

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$v_{fy}=2.63m/s$

Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

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