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Vesnalui [34]
4 years ago
9

Suppose you are in an elevator that is moving upward with a constant velocity. A scale inside the elevator shows your weight to

be 600 N.
(a) Does the scale register a value that is greater than, less than, or equal to 600 N during the time when the elevator slows down as it comes to a stop?
(b) What is the reading when the elevator is stopped?
(c) How does the value registered on the scale compare to 600 N during the time when the elevator picks up speed again on its way back down? Give your reasoning
Physics
1 answer:
Ulleksa [173]4 years ago
4 0

Answer:

a) Less than 600 N

b) 600 N

c) Less than 600 N

Explanation:

The value registered by the scale, measures the normal force on the object placed on the scale.

The person on the scale, is acted by two external forces, gravity (which we call weight, and always go downward) and the normal force, an upward force in this case.

The difference between both forces, according Newton’s 2nd law, must be equal to the product of the mass of the object, times the acceleration.

If the elevator moves upward to a constant speed, and then slows down, this means that there exists a net force on the person, producing a acceleration directed downwards on him.

If we take the direction of the acceleration (downward) to be positive, this means that the difference between gravity force and normal force must be positive.

As mg = 600 N (equal to normal force when no net force is present), the normal force in this case must be less than 600 N.

b) If the elevator is stopped, the effect is the same like it were moving at constant speed, so in this case the normal force remains the same: 600 N

c) Assuming that it starts from rest, if it accelerates going downward, and if we take as positive the downward direction, in order to satisfy Newton’s 2nd Law, normal force must be less than 600N so the difference between gravity and normal force remain positive.

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Please help!!
pantera1 [17]

Answer:

(a) The time the ball stays in the air is approximately 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is approximately 83.18 meters

(c) The maximum altitude of the ball is approximately 26.62 m

Explanation:

The given parameters of the question are;

The initial velocity at which a baseball cannon fires balls, u = 29 m/s

The angle at which the cannon is tilted, θ = 52°

(a) The time duration the ball stays in the air is given by the time of flight of the projected ball as follows;

2 \cdot t = \dfrac{2 \cdot u \cdot sin (\theta)}{g}

Where;

t = The time it takes the baseball to reach maximum height

2·t = The total time of flight = The time the ball stays in the air

θ = The angle at which the ball is tilted = 52°

g = The acceleration due to gravity ≈ 9.81 m/s²

u = The initial velocity = 29 m/s

Therefore, we have;

2 \cdot t = \dfrac{2 \times 29 \ m/s\times sin (52^{\circ})}{9.81 \ m/s^2} \approx 4.66 \ s

The time the ball stays in the air, 2·t ≈ 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is given by the horizontal range, 'R', of the projectile as follows;

Horizontal \ range, R = \dfrac{u^2 \cdot sin(2 \cdot \theta) }{g}

∴ The distance from the cannon at which the ball will hit the ground = R

Horizontal \ range, R = \dfrac{(29 \ (m/s))^2 \cdot sin(2 \times 52^{\circ}) }{9.81 \ m/s^2} \approx 83.18 \, m

The distance from the cannon at which the ball will hit the ground = R ≈ 83.18 meters

(c) The maximum altitude of the ball is equal to the maximum height reached by the ball, H, which is given as follows;

H = \dfrac{u^2 \cdot sin^2 (\theta)}{2 \cdot g}

Therefore, we have;

H = \dfrac{(29 \ m/s)^2 \times sin^2 (52^{\circ})}{2 \times 9.81 \ m/s^2} \approx 26.62 \ m

The maximum altitude of the ball ≈ 26.62 m

3 0
3 years ago
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