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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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Alik [6]

Answer:

The correct answer to the following question will be Option A (moment arm; pivot point).

Explanation:

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  • The moment arm extends this same distance from either the pivot point to just the position of that same pressure exerted.
  • The pivotal point seems to be the technical indicators required to fully measure the appropriate demand trends alongside different time-frames.

The other three choices are not related to the given situation. So that option A is the appropriate choice.

7 0
2 years ago
Seema knows the mass of basketball. What other information is needed to find the balls potential energy
Lelu [443]

Answer: The height (position) of the ball and the acceleration due gravity

Explanation:

In this case we are taking about gravitational potential energy, which is the energy a body or object possesses, due to its position in a gravitational field.  In this sense, this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.  

In the case of the Earth, in which the gravitational field is considered constant, the gravitational potential energy U will be:  

U=mgh  

Where:

m is the mass of the ball

g=9.8 m/s^{2} is the acceleration due gravity (assuming the ball is on the Earth surface)

h is the height (position) of the ball respect to a given point

Note the value of the gravitational potential energy is directly proportional to the height.

8 0
2 years ago
Read 2 more answers
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes
OlgaM077 [116]

Answer:

Explanation:

mass of car, m = 1000 kg

initial velocity, u = 20 m/s

final velocity, v = 0 m/s

distance, s = 120 m

Let a be the acceleration of motion

use third equation of motion

v² = u² + 2 as

0 = 20 x 20 + 2 x a x 120

a = - 1.67 m/s²

Let F be the force

Force, F  mass x acceleration

F = - 1000 x 1.67

F = - 1666.67 N

The direction of force is towards south and the magnitude of force is 1666.67 N.

8 0
2 years ago
Which of the following statements is FALSE?
Levart [38]

A thrust fault is a reverse fault with an extremely high dip (close to 90°). This is the false statement.

Answer: Option D

<u>Explanation:</u>

Faults are the fracture or fracture zone occurring on the rocks. These fractures can travel through the rocks leading to massive destruction. So, depending upon the direction of their travel, the faults can be classified as normal, reverse and strike slip fault. Also, the angle of dip along the fault is one of the important criteria for determining the type of faults.

There is dip-slip fault which has its movement along the vertical fault plane while the strike slip fault will be in horizontal direction. Similarly, an oblique fault will be acting in both vertical and the horizontal direction. So, the fourth statement related to thrust fault is false as in reverse fault or thrust fault the dip will be shallow and not high.

5 0
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A cyclist moving with a constant velocity of 6.0 m/s forward passes a car that is just starting. If the car has a constant accel
sergiy2304 [10]

After 6 seconds, the car will surpass the cyclist.

<h3><u>Explanation:</u></h3>

The speed of the cyclist = 6 m/s.

Let after time t sec, the car will overtake the cyclist.

So, distance covered by the cyclist in t sec = 6t m

Initial velocity of the car is 0 m/s, because the car is just starting.

Acceleration of the car =2 m/s^2.

Final velocity of the car =6 m/s.

So to cover the distance 6t, the time required by the car = \frac{1}{2} \times a \times t^2 = \frac{1}{2} \times 2 \times t^2

6t = \frac{1}{2} \times 2 \times t^2\\6t = t^2

t =6 sec

So, after 6 seconds, the car will surpass the cycle.  

6 0
2 years ago
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