When a body performs a uniform circular motion, the direction of the velocity vector changes at every moment. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circle that gives rise to centripetal acceleration, the mathematical expression is given as,
Where,
v = Tangential Velocity
r = Radius
The linear velocity was 2010m/s in a radius of 0.159m, then the centripetal acceleration is
Therefore the centripetal acceleration of the end of the rod is 
1 volt = 1 joule per coulomb
It takes 1 joule of work to force a coulomb of charge enough closer to a charge
with the same sign to raise its potential 1 volt.
If you allow 1 coulomb of charge to fall to where its potential is 1 volt less,
it gives up 1 joule of energy.
A lot. A lot of bytes.
But I suppose you're looking for a more specific answer, huh?
<span>1 Gigabyte = 1,073,741,824 Bytes
</span>
That external computer flash drive can hold 1,073,741,824 bytes of data.
Answer:
(a) 333.77 J
(b) 237.85 J
(c) 4763.77 J
(d) 4667.85 J
Explanation:
Temperature of source, TH = 314 K
Temperature of A, Tc = 292 K
Temperature of B, Tc' = 298 K
heat taken out, Qc = 4430 J
Let the heat deposited outside is QH and QH' by A and B respectively.
Now
(a) Work done for A
W = QH - QC = 4763.77 - 4430 = 333.77 J
(b) Work done for B
W' = QH' - Qc = 4667.85 - 4430 = 237.85 J
(c) QH = 4763.77 J
(d) QH' = 4667.85 J
Using newton's law of gravity on a 1kg mass near the star. G is newton's grav constant. g is acceleration of grav at starGMstar/r^2 = gstar
