The kinetic energy of an object of mass m and velocity v is given by

Let's call

the initial speed of the car, so that its initial kinetic energy is

where m is the mass of the car.
The problem says that the car speeds up until its velocity is twice the original one, so

and by using the new velocity we can calculate the final kinetic energy of the car

so, if the velocity of the car is doubled, the new kinetic energy is 4 times the initial kinetic energy.
Answer: False
Explanation:
Relative to the concept of radiations, a black body is an object capable of absorbing any form of electromagnetic radiation irrespective of its frequency or angle of incidence when incident on such object.
However, the same cannot be said about real bodies as real bodies are those which reflect all rays incident on them completely and uniformly in all directions.
One very important characteristic of black bodies is that they are ideal emmiters.
The concept of emmisivity is brought about by the existence of real bodies .
This is due to the fact that they are only able to emit radiation at a fraction of the black body energy levels.
Please note that by convention, the emmisivity of a real body is always less thaan 1.
As such they are not able to emit as much radiation as a black body at the same temperature.
Answer: I think the answer is a wave/ a transverse wave, or electromagnetic waves as well.
Actual displacement that he required to move
towards North
Displacement that he moved due to snow is
at 47 degree North of East
now in vector component form we can say



now the displacement that is more required to reach the destination is given as



so the magnitude of the displacement is given as


its direction is given as

so it is 5.54 km towards 5.38 degree North of West.