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3 years ago

A 57 kg woman is on an elevator accelerating upward at "1.25" meters per second squared. what is the normal force acting on her?

Physics

1 answer:

ella [17]3 years ago

8 0

Answer:

Normal force acting on the woman is 629.85 N.

Explanation:

Given:

Mass of the woman is, $m=57\textrm{ kg}$

Net upward acceleration is, $a=1.25\textrm{ }m/s^{2}$

Acceleration due to gravity, $g=9.8\textrm{ }m/s^{2}$

Let the normal force acting upward be $R$ newtons.

Therefore, net force in the upward direction is given as:

Net force = Upward force - Downward force.

Downward force acting on the woman is her weight which is equal to $mg$ .

Therefore, Net force = $R-mg$

Now, as per Newton's second law of motion,

Net force, $F_{net}=ma$

So,

$R-mg=ma\\R=mg+ma\\R=m(g+a)$

Plug in 57 kg for $m$ , 9.8 m/s² for $g$ , and 1.25 m/s² for $a$ . Solve for $R$ . This gives,

$R=57(9.8+1.25)\\R=57(11.05)=629.85\textrm{ N}$

Therefore, the normal force acting on her is 629.85 N.

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Answer:

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3 years ago

A car that is standing still accelerates up a hill with a slope of 6.4% at an average acceleration of 2.93 ft/s^2. The hill is 1

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Answer:

213 s

Explanation:

Slope is the ratio of change in vertical distance to change in horizontal distance.

Slope = vertical height / horizontal height

Therefore:

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3 years ago

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Answer:

c. about 1/10 as great.

Explanation:

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