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
<span>Kinetic energy is energy that a body possesses by virtue of being in motion.
Think of a windmill...
Windmills use the winds energy in order to spin and in doing so creates energy for other things to be powered.
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A car acting as an object in front of a biconvex lens between F and 2 F on the object side of the lens. There is a light ray parallel to the principal axis that is bent through F on the image side of the lens. There is a ray straight through the center of the lens. The rays intersect below the x axis further than 2 F away from the lens and farther from the principal axis than the object is tall.
<u> The image produced by the lens is (b) inverted and real</u>
Explanation:
A real image occurs where the rays converge.
Real images can be produced both by the concave mirrors or converging lenses, but the condition is that the object of consideration is always placed far away from the mirror or the lens than the focal point, and thus the real image produced is inverted.
A car acting as an object in front of a biconvex lens between F and 2 F on the object side of the lens. There is a light ray parallel to the principal axis that is bent through F on the image side of the lens. There is a ray straight through the center of the lens. The rays intersect below the x axis further than 2 F away from the lens and farther from the principal axis than the object is tall.
<u> The image produced by the lens is (b) inverted and real</u>
(a) 
First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:
where
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 1.00 cm = 0.01 m is the displacement of the person
Solving for a,
And the average force on the person is given by
with m = 75.0 kg being the mass of the person. Substituting,
where the negative sign means the force is opposite to the direction of motion of the person.
b) 
In this case,
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 15.00 cm = 0.15 m is the displacement of the person with the air bag
So the acceleration is
So the average force on the person is
Answer:
what is that supposed to even mean
Explanation:
