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3 years ago

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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Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms

amid [387]

Answer:

a) Work required for air conditioner A = 354.7 J

b) Work required for air conditioner B = 310.3 J

c) The magnitude of the heat deposited outside for conditioner A = 4684.7 J

d) The magnitude of the heat deposited outside for conditioner B = 4640.3 J

Explanation:

In a carnot air conditioner, it operates like a reverse carnot engine; i.e. it removes heat from the cold reservoir (making it colder) and dumps the heat in the hot reservoir (making it hotter)

For a Carnot air conditioner,

Q꜀ is the heat removed from the colder reservoir = 4330 J for both cases

T꜀ is the temperature of the colder reservoir (temperature of the rooms) = 293 K and 296 K for both cases to be considered.

Qₕ is the heat deposited in the warmer reservoir = ? for both cases

Tₕ is the temperature of the hot reservoir (temperature of outside) = 317 K for both cases.

For Carnot air conditioners,

Qₕ = W + Q꜀ (eqn 1)

And

(Qₕ/Tₕ) - (Q꜀/T꜀) = 0 (eqn 2)

Making Qₕ the subject of formula in eqn 2

Qₕ = Tₕ (Q꜀/T꜀)

Substituting this into eqn 1

Tₕ (Q꜀/T꜀) = W + Q꜀

Q꜀ (Tₕ/T꜀) - Q꜀ = W

Q꜀ [(Tₕ - T꜀)/T꜀ ] = W

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ]

For the air conditioner A,

T꜀ = 293 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 293)/293] = 354.7 J

For the air conditioner B,

T꜀ = 296 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 296)/296] = 310.3 J

c) Qₕ = W + Q꜀

For conditioner A,

Qₕ = 354.7 + 4330 = 4684.7 J

For conditioner B,

Qₕ = 310.3 + 4330 = 4640.3 J

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