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3 years ago

A person of weight mg, is standing on a scale in an elevator. What will the

Physics

1 answer:

sergejj [24]3 years ago

8 0

Answer:

B. less than mg

Explanation:

A person's weight changes as he rides an elevator at a constant velocity. If the elevator is moving downward, the person's weight becomes lesser. He is considered to be in "free fall" due to gravity. However, the opposite happens when the elevator goes upward. The person will feel heavier because he is going against gravity. This means his weight is more than his actual weight. If the person will stand on an elevator that doesn't move, his weight will just be the same. All of these is due to Newton's Second Law of Motion where an object's acceleration relies on its force and mass.

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4 years ago

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3 years ago

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4 years ago

What could be the possible answer to the question ? thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately 0.377·M·g

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$

F₀ ≈ 0.377·M·g

Learn more about equilibrium of forces here:

brainly.com/question/6995192

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2 years ago

Read 2 more answers