No. I Is actually V/R. R/V would be 1/I.
Answer: Impulse = 4 kgm/s
Explanation:
From the question, you're given the following parameters:
Momentum P1 = 12 kgm/s
Momentum P2 = 16 kgm/s
Time t = 0.2 s
According to second law of motion,
Force F = change in momentum ÷ time
That is
F = (P2 - P1)/t
Cross multiply
Ft = P2 - P1
Where Ft = impulse
Substitute P1 and P2 into the formula
Impulse = 16 - 12 = 4 kgm/s
The magnitude of the impulse is therefore 4 kgm/s.
Answer:
Explanation:
a )
change in the gravitational potential energy of the bear-Earth system during the slide = mgh
= 45 x 9.8 x 11
= 4851 J
b )
kinetic energy of bear just before hitting the ground
= 1/2 m v²
= .5 x 45 x 5.8²
= 756.9 J
c ) If the average frictional force that acts on the sliding bear be F
negative work done by friction
= F x 11 J
then ,
4851 J - F x 11 = 756.9 J
F x 11 = 4851 J - 756.9 J
= 4094.1 J
F = 4094.1 / 11
= 372.2 N
Explanation:
Since, it is mentioned the there occurs no change in the temperature. This also means that there will occur no change in thermal energy of the system.
Hence, = 0. And, as = 0 then there will be no work involved. This means that total energy added to the house will return to the outside air as heat.
Therefore,
Q = -(19000 J + 2000 J)
= -21000 J
or, |Q| = 21000 J
Thus, we can conclude that the magnitude of the energy transfer between the house and the outside air is 21000 J.
I guess you could say both answers are true in the context you take.