Answer:
a) Instantaneous rate at which the force does work on the object = -6 W
b) 
Explanation:
a) Given that

and

Instantaneous rate at which the force does work on the object is called power.
Power is the dot product of force and velocity.

Instantaneous rate at which the force does work on the object = -6 W
b) Here given that 
Power = -12 W


Answer:
0.0192A
Explanation:
Since, the reading of the galvanometer is 0 A, the voltage across the resistance R will be:
Step 1
VR = V2
VR = 3.00v
Step 2
Calculating the current through the resistance R as below,
IR = V1 - V2 /R1
IR = 12 - 3 /468
IR =0.0192A
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