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Artemon [7]
3 years ago
14

A 38.5 kg man is in an elevator that is moving at a constant speed. How much normal force is being exerted on him?

Physics
1 answer:
Alexus [3.1K]3 years ago
6 0

Answer:

Force = mass times acceleration

Acceleration is change in velocity over time.

The elevator is moving at constant velocity. That means there is no net force on the elevator. Gravity is still acting on the elevator but the tension of cables, friction and other mechanism are cancelling out the acceleration due to gravity. The same as an elevator being held at a floor.

The normal force is the force exerted by surfaces the to stop you from passing through. It is perpendicular to the surface the object is on, in this case the 38.5kg person. It being an elevator I am assuming the floor is perpendicular to the force of gravity. Normal force is equal to the force being exerted on it, in this scenario the persons mass x acceleration due to gravity.

So your answer is 38.5kg x 9.81m/s^2 = 377.685N

Explanation:

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3 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
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Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

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Answer:

unmmmmmmmm I think the answerA

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