Answer:
6692J
Explanation:
Power is defined as the rate at which work is being done.
So,
Power = 
Work done = Power x time
Given parameters:
Power = 478watts
Time = 14s
So;
Work done = 478 x 14 = 6692J
"Balanced" means that if there's something pulling one way, then there's also
something else pulling the other way.
-- If there's a kid sitting on one end of a see-saw, and another one with the
same weight sitting on the other end, then the see-saw is balanced, and
neither end goes up or down. It's just as if there's nobody sitting on it.
-- If there's a tug-of-war going on, and there are 300 freshmen pulling on one
end of a rope, and another 300 freshmen pulling in the opposite direction on
the other end of the rope, then the hanky hanging from the middle of the rope
doesn't move. The pulls on the rope are balanced, and it's just as if nobody
is pulling on it at all.
-- If a lady in the supermarket is pushing her shopping cart up the aisle, and her
two little kids are in front of the cart pushing it in the other direction, backwards,
toward her. If the kids are strong enough, then the forces on the cart can be
balanced. Then the cart doesn't move at all, and it's just as if nobody is pushing
on it at all.
From these examples, you can see a few things:
-- There's no such thing as "a balanced force" or "an unbalanced force".
It's a <em><u>group</u> of forces</em> that is either balanced or unbalanced.
-- The group of forces is balanced if their strengths and directions are
just right so that each force is canceled out by one or more of the others.
-- When the group of forces on an object is balanced, then the effect on the
object is just as if there were no force on it at all.
Answer:
B
Explanation:
The control is something that is meant to not be changed, the control is a comparison of the experimental.
Well the trivial answer is zero, since there is indeed a "zero vector". Assuming you aren't allowed to use the zero vector you would need at least two. They would be antiparallel and of equal magnitude. (That is be pointing in opposite directions and have the same length)
Answer:
Explanation:
Given that,
Mass of the heavier car m_1 = 1750 kg
Mass of the lighter car m_2 = 1350 kg
The speed of the lighter car just after collision can be represented as follows
b) the change in the combined kinetic energy of the two-car system during this collision
substitute the value in the equation above
Hence, the change in combine kinetic energy is -2534.78J
