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2 years ago

6

A force of 120 N is exerted on a 40 kg container which sits on a floor. If the frictional force between floor and container is 8

0 N, what is the magnitude of the acceleration of the container?

1 answer:

harkovskaia [24]2 years ago

7 0

Answer:

Magnitude is 144N .

Explanation:

a²+b²=c²

120N²+80N²=c²

14400+6400=c²

$\sqrt{20800} = \sqrt{c} ^{2}$

144N=c

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rewona [7]

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Explanation:

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By the ideal gases law: PV=RTn

Then, in (1): (remember Pa = Pb)

W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)

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