Answer:
194,400 joules of kinetic energy.
Explanation:
Remember that to calculate the Kinetic energy you need to use the next formula:

We know that Mass= 1200 kg and velocity is 18m/s, so we insert those values into the formula:

So the kinetic energy of a car moving at 18m/s with a mass of 1200 kg would be 194,400 joules.
Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion

Here s = h,u = 450m/s a = -g and t = t+3
Substituting

Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting

Solving both equations

So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s

Passing of B occurs at 4108.31 height.
The vector, the x-component and the y-component form a rectangle triangle where the vector is the hypothenuse and the x and y components are the two sides.
Calling

the angle between the vector and the horizontal direction (x), the two sides are related to

by

where vy and vx are the two components on the y- and x-axis. Using vx=10 and vy=3 we find

And so the angle is
Answer:
(a) 10 m/s
(b) 22.4 m/s
Explanation:
(a) Draw a free body diagram of the car when it is at the top of the loop. There are two forces: weight force mg pulling down, and normal force N pushing down.
Sum of forces in the centripetal direction (towards the center):
∑F = ma
mg + N = mv²/r
At minimum speed, the normal force is 0.
mg = mv²/r
g = v²/r
v = √(gr)
v = √(10 m/s² × 10.0 m)
v = 10 m/s
(b) Energy is conserved.
Initial kinetic energy + initial potential energy = final kinetic energy
½ mv₀² + mgh = ½ mv²
v₀² + 2gh = v²
(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²
v = 22.4 m/s
Earth is 150 million kilometers away for the sun