Well, there's a lot of friction going on there, so the snowball gradually
loses kinetic energy just from bouncing and plowing through the snow
on the ground.
But I don't think you're asking about that. I think you're ignoring that
for the moment, and asking how its kinetic energy changes as its
mass increases. We know that
Kinetic Energy = (1/2) (mass) (speed²)
and THAT seems to say that more mass means more kinetic energy.
So maybe the snowball's kinetic energy increases as it picks up
more mass.
Don't you believe it !
Remember: Energy always has to come from somewhere ... a motor,
a jet, a push, gravity ... something ! It doesn't just appear out of thin air.
If the snowball were rolling down hill, then it could get more kinetic energy
from gravity. But if it's rolling on level ground, then it can never have any
more kinetic energy than you gave it when you pushed it and let it go.
If snow or leaves stick to it and its mass increases, then its speed must
decrease, in order to keep the same kinetic energy.
i just had this question and i totally guessed. it was 2.2.
Imagine you were able to throw a ball in a frictionless environment
such as outer space. Once you let go of the ball, it will travel forever
in a straight line, and at a constant speed. (At least until it bumps into
something.)
A car accelerates down the road. The reaction to the tires pushing
on the road is the road pushing on the tires.
Answer: A. 200J
Therefore, the workdone by the heat engine is 200J
Explanation:
Given ;
The efficiency of the heat engine is E = 20% = 0.2
Heat loss L= 800J
For an heat engine the efficiency is measured by the amount of workdone by the heat engine when compared to the heat generated.
Efficiency E = workdone/heat generated × 100%
Heat generated G= workdone W + heat loss L
G = W + L
According to the question.
W = 20% of G
W = 0.2G ......1
L = 80% of G
L = 0.8G
G = L/0.8 ......2
Substituting equation 2 to 1
W = 0.2(L/0.8)
And L = 800J
W = 0.2(800/0.8)
W = 200J
Therefore, the workdone by the heat engine is 200J
