The work done onto the car is 506,250 J
The work done on a system implies an increase in the internal energy of the system as a result of some forces acting on the system from the outside.
From the parameters given:
- The mass of the car = 1500 kg
- The initial speed = 30 m/s
- The final speed = 15 m/s
The work done onto the car refers to the change in the kinetic energy (i.e. ΔK.E)
= 506,250 J
Therefore, we can conclude that the work done on the car is 506,250 J
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Answer:
7.3cm above the compressed spring.
Explanation:
We can use the conservation energy theorem to solve this problem:
The block was dropped 7.3cm above the compressed spring.
<u>Answer:</u>
Option: D. Gravity is pulling the crash test dummy in the direction the car is moving.
<u>Explanation:
</u>
When a car accelerates from a standing start, the crash test dummy appears to be pressed backward into the seat cushion because the gravity is pulling the crash test dummy in the direction the car is moving.
Basically when the car is starting, the person inside is in static position and the car is going to move. So it is putting a force on the person to move on the same speed. But as the person is sitting static hence gravity is pulling him behind from moving. Hence, The dummy appears to be pressed backward.
It would be helpful if you draw the figure of the problem. You will see that a right triangle would be constructed by the problem where 19.0 is the angle between the hypotenuse and the base of the triangle. It is said that the force acting is said to be 9.0 N at the said angle to the horizontal. Using trigonometric relations,
cos 19 = adjacent / hypotenuse = horizontal component / 9
horizontal component = 8.51 N
Answer:
A relationship where both members are freindly and supportive towards each other.
Explanation:
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