Answer:
It must be 4 times high.
Explanation:
- Assuming that the car can be treated as a point mass, and that the ramp is frictionless, the total mechanical energy must be conserved.
- This means, that at any time, the following must be true:
- ΔK (change in kinetic energy) = ΔU (change in gravitational potential energy)
⇒
- Let's call v₁, to the final speed of the car, and h₁ to the height of the ramp.
So, at the bottom of the ramp, all the gravitational potential energy
must be equal to the kinetic energy of the car (Defining the bottom of
the ramp as our zero reference for the gravitational potential energy):
(1)
- Now, let's do v₂ = 2* v₁
- Replacing in (1) we get:
(2)
- Dividing (2) by (1), and rearranging terms, we get:
- h₂ = 4* h₁
In this case the rubber raft has horizontal and vertical motion.
Considering vertical motion first.
We have displacement , u = Initial velocity, t = time taken, a = acceleration.
In vertical motion
s = 1960 m, u = 0 m/s, a = 9.81
So raft will take 20 seconds to reach ground.
Now considering horizontal motion of raft
u = 109 m/s, t = 20 s, a = 0
So
So shipwreck was 2180 meter far away from the plane when the raft was dropped.
Answer:
m = 9795.9 kg
Explanation:
v = 35 m/s
KE = 6,000,000 J
Plug those values into the following equation:
6,000,000 J = (1/2)(35^2)m
---> m = 9795.9 kg
This is false. they flow west to east