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Afina-wow [57]
2 years ago
12

For each of the motions described below, determine the algebraic sign (+, -, or 0) of the velocity and acceleration of the objec

t at the time specified. For all of the motions, the positive y axis is upward.Part A. An elevator is moving downward when someone presses the emergency stop button. The elevator comes to rest a short time later. Give the signs for the velocity and the acceleration of the elevator after the button has been pressed but before the elevator has stopped.Part B. A child throws a baseball directly upward. What are the signs of the velocity and acceleration of the ball immediately after the ball leaves the child's hand?Part C. A child throws a baseball directly upward. What are the signs of the velocity and acceleration of the ball at the very top of the ball's motion (i.e., the point of maximum height)?
Physics
1 answer:
Wittaler [7]2 years ago
7 0

Answer:

A. Velocity is negative (-)

Acceleration is positive,(+)

B. Velocity is positive. (+)

Acceleration is negative (-)  

C. Velocity is zero (0).

Acceleration is negative (-)

Explanation:

The elevator is said to be moving downward, therefore, its motion is in the negative direction as the positive direction is upward in the y-axis. Velocity is negative (-)

As the elevator is making an emergency stop, it is decelerating. Deceleration is negative acceleration. However, since it occurs in the opposite direction, i.e. acceleration vector is pointing upward, acceleration is positive,(+)

The motion of the ball is in the upward direction, therefore the velocity is positive. (+)

The acceleration due to the force of gravity acts in the opposite direction to that of the ball, i.e. downwards, acceleration is negative (-)  

At maximum height, the ball will stop moving, therefore, velocity is zero (0).

Since acceleration due to the force of gravity acts downward, the acceleration is negative (-)

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Explanation:

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Assuming that,

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Using Bernoulli equation

P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''

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Then, Bernoulli equation becomes

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Rearranging

P' — P'' = ½pv"² —½pv'²

P'—P" = ½p ( v"² —v'²)

P'—P" = ½ × 1.29 × (66²-40²)

P'—P" = 1777.62 N/m²

Lift force can be found from

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Force = ∆P ×A

Force = (P' —P")×A

Since we already have (P'—P")

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W = 1777.62 × 17

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Given 75-kg sprinter accelerates from rest to a speed of 11.0 m/s in 5.0 s.

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