Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s
Answer:

Explanation:
Set up force equation
∑F=ma
∑F=W+FB
The minus sign for downward direction
What type of change occurs when water changes from solid to a liquid a phase change a physical change and irreversible change both A and the
Answer:

Explanation:
Work is the product of force and distance.

We know the force to lift the rock was 50 Newtons. The work was 150 Joules.
- 1 Joule is equal to 1 Newton meters.
- We can convert the units to make the problem simpler later. The work is also 150 Newton meters.

Substitute the values into the formula.

We want to solve for distance, so we must isolate the variable. Divide both sides of the equation by 50 Newtons.

The Newtons will cancel out.

The rock was lifted <u>3 meters.</u>
From the equation;
a= -1÷2gT2
a= -5×3(3)
a= -45m/s
it will drop about -45m/s2.
Answer:
15"
Explanation:
Let x be the length of the longer part, then the length of the shorter part is 3x/5, or also x - 6
therefore we have the following equation:
3x/5 = x - 6
We can multiply both sides by 5
3x = 5x - 6*5
2x = 30
x = 30/2 = 15"
So the length of the longer part is 15"