This question involves the concepts of the equations of motion, kinetic energy, and potential energy.
a. The kinetic energy of the rocket at launch is "3.6 J".
b. maximum gravitational potential energy of the rocket is "3.6 J".
<h3>a. KINETIC ENERGY AT LAUNCH</h3>
The kinetic energy of the rocket at launch is given by the following formula:

where,
- K.E = initial kinetic energy = ?
- m = mass of rocket = 0.05 kg
= initial speed = 12 m/s
Therefore,

K.E = 3.6 J
<h3>
b. MAXIMUM GRAVITATIONAL POTENTIAL ENERGY</h3>
First, we will use the third equation of motion to find the maximum height reached by rocket:

where,
- g = -9.81 m/s²
- h = maximum height = ?
- vf = final speed = 0 m/s
Therefore,
2(-9.81 m/s²)h = (0 m/s)² - (12 m/s)²
h = 7.34 m
Hence, the maximum gravitational potential energy will be:
P.E = mgh
P.E = (0.05 kg)(9.81 m/s²)(7.34 m)
P.E = 3.6 J
Learn more about the equations of motion here:
brainly.com/question/5955789
Answer:
Maximum force, F = 1809.55 N
Explanation:
Given that,
Diameter of the anterior cruciate ligament, d = 4.8 mm
Radius, r = 2.4 mm
The tensile strength of the anterior cruciate ligament, 
We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,

So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N
Answer:
Explanation:
For starters begin with a warning not to touch the brake drums. All of the KE is transferred to the brake drums. The result is a large rise in temperature. Heat. If you press hard on the brakes, rubber is left on the road and there is heat involved in that too.