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tekilochka [14]
2 years ago
14

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.
Enter the correct sign for the elevator's velocity and the correct sign for the elevator's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,- . If you think that both are zero, then you would enter 0,0 .


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?
Enter the correct sign for the baseball's velocity and the correct sign for the baseball's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,- . If you think that both are zero, then you would enter 0,0 .


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)?
Enter the correct sign for the baseball's velocity and the correct sign for the baseball's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,- . If you think that both are zero, then you would enter 0,0 .
Physics
1 answer:
evablogger [386]2 years ago
7 0

Part A: -,+

The elevator is moving downward, this is what determines the direction of the velocity, as it will follow the direction of the movement. As we are told that the positive direction is upward, then the velocity has negative direction. Also, after the button is pressed, the elevator starts to stop, in other words, its velocity starts to decreased. This means that the acceleration has an opposite direction to the velocity, therefore, its sign is +.

Part B: +, -

The ball is moving upward, and as said before, this is what determines the direction of the velocity, as it will follow the direction of the movement. Then, velocity has a + sign.

Also, after the ball is thrown, there is no other force other than gravity, which will oppose to the movement of the ball, trying to make it come back to the ground. This means that the acceleration has an opposite direction to the velocity, in other words, it's directed downward, therefore, its sign is -.

Part C: 0, -

The acceleration of the ball since it was thrown until it fell to the ground will always be the gravity, which will always go downward (-).

After being thrown, the ball's velocity will start to decrease because of gravity. When its velocity has turned to 0, the ball will have reached maximum height . At this point it will start to fall again, accelerated by gravity. But at the very top, the velocity of the ball is 0.

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Answer:

980 J

Explanation:

The change in box's energy is equal to its change in gravitational potential energy:

\Delta U = m g \Delta h

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m = 50 kg is the mass of the box

g = 9.8 m/s^2 is the acceleration due to gravity

\Delta h= 2m is the change in height of the box

Substituting numbers, we find

\Delta U = (50 kg)(9.8 m/s^2)(2 m)=980 J

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An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start
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Answer:

  • The total distance traveled is 28 inches.
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Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector \hat{i} pointing in the west direction, the ant start at position

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so, the distance traveled here is

d_1 = |\vec{r}_1 - \vec{r}_0  | = | 29 \ inch   \ \hat{i} - 16 \ inch   \ \hat{i}  |

d_1 =  | 13 \ inch   \ \hat{i}  |

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after this, the ant travels to

\vec{r}_2 = 14 \ inch \ \hat{i}

so, the distance traveled here is

d_2 = |\vec{r}_2 - \vec{r}_1  | = | 14 \ inch   \ \hat{i} - 29 \ inch   \ \hat{i}  |

d_2 =  | - 15 \ inch   \ \hat{i}  |

d_2 =  15 \ inch

The total distance traveled will be:

d_1 + d_2 = 13 \ inch + 15 \ inch = 28 \ inch

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