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Mazyrski [523]
3 years ago
7

A roller coaster car has 600,00 J of Kinetic energy as it approaches the station to stop. The roller coaster comes to a complete

stop. How much work did the brakes do in stopping the car?
Physics
1 answer:
guajiro [1.7K]3 years ago
7 0

Answer:

Work Done by the Brakes = - 60000 J

Negative sign indicates the opposite direction of force and displacement.

Explanation:

Assuming that frictional effects are negligible, we can say that all the work done by the brakes is used to stop the car. We can apply law of conservation of energy to this situation as follows:

Work Done by the Brakes = Change in Kinetic Energy

Work Done by the Brakes = Final Kinetic Energy - Initial Kinetic Energy

Final Kinetic Energy of the roller coaster will be 0. Because, the roller coaster will stop finally and its velocity will become zero.

Work Done by the Brakes = 0 J - 60000 J

<u>Work Done by the Brakes = - 60000 J</u>

<u>Negative sign indicates the opposite direction of force and displacement.</u>

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a ball is thrown down at 25 m/s from a 500m tall building. how fast is it traveling when it hits the ground?
Mrrafil [7]

Answer:

The speed of the ball when it hits the ground is 102.1 m/s

Explanation:

Given;

initial velocity of ball, u = 25 m/s

distance traveled by the ball = height of the building = h = 500 m

when the ball hits the ground, the final velocity, v = ?

The final velocity of the ball is given by;

v² = u² + 2gh

where;

g is acceleration due to gravity = 9.8 m/s²

v² = (25)² + 2(9.8)(500)

v² = 10425

v = √10425

v = 102.1 m/s

Therefore, the speed of the ball when it hits the ground is 102.1 m/s

3 0
3 years ago
A 50 kilogram woman wearing a seatbelt is traveling in a car that is moving with a velocity of +10 meters per second. In an emer
ra1l [238]

Answer:

the answer is A.) -1 * 10^3[N]

Explanation:

The solution consists of two steps, the first step is using the following kinematic equation:

v=v_{i} +a*t\\where:\\v=final velocity [m/s]\\v_{i}=initial velocity [m/s]\\a=acceleration[m/^2]\\t=time[s]\\

The initial velocity is 10 [m/s], and the final velocity is zero because the car stops in 0.5[s].

Replacing:

0=10+a*(0.5)\\a=-20[m/s^2]

Now in the second part, we need to use the second law of Newton, this law relates the forces with the acceleration of a body.

In the moment when the car stops suddenly the driver will feel the force of the seatbelt acting in the opposite direction of the movement.

F=m*a\\F=50[kg]*(-20[m/s^2])\\units\[kg]*[m/s^2]=[N]\\F=-1000[N] or -1*10^{3} [N]

The minus sign means that the force is acting in the opposite direction of the movement.

7 0
3 years ago
Can someone please give me the answers to this? ... please ...
alexira [117]
3. 2 meters per second 4. i think object 7. i’ll try to figure it out 8. .77 9. 25km 10. 10m each second
7 0
3 years ago
A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is paralle
Alex73 [517]

Answer:

715 N

Explanation:

Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.

Let g = 9.8 m/s2

Gravity and equalized normal force is:

N = P = mg = 107*9.8 = 1048.6 N

Kinetic friction force and equalized tension force on the rope is

T = F_{\mu} = N\mu = 1048.6 * 0.682 = 715.1452 N

6 0
3 years ago
At a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienc
vaieri [72.5K]

Answer:

27.1 m/s

Explanation:

Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.

Using third equation of motion,

V^2 = U^2 + 2aS

Since the car is decelerating, the final velocity V = 0

Substitute all the parameter into the equation above,

0 = U^2 - 2 * 40.52 * 9.06

U^2 = 734.22

U = \sqrt{734.22}

U = 27.096

U = 27.1 m/s  approximately

Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid

5 0
3 years ago
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