Heat = change in internal energy + Work done The internal energy of a system = heat added and mechanical work done by the system, i.e. U = Q + W rearranging the formula above, will give us: Q = deltaU + W
Q = U - W = 604 kJ - 43.0 kJ = 561,000 J would be the answer.
Answer:
Same magnitude of the 10 nc charge cause the electric field is external.
Explanation:
To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.
As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.
F = qE
If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.
Answer:
Explanation:
given,
Angular speed of the tire = 32 rad/s
Displacement of the wheel = 3.5 rev
Δ θ = 3.5 x 2 π
= 7 π rad
now,
Time interval of the car to rotate 7π rad
using equation
Time taken to rotate 3.5 times is equal to 0.687 s.
Tsunami? I think so maybe it’s right
Its average speed, pretending that it traveled at a constant speed, is
v = s / t
= 600 m
5 x 60 s
= 2 m/s
but to be a velocity it needs a direction as well as a speed.
( Sorry. Can’t find a division line to put between the 600 m and the 5 x 60 s )
