Answer:
27 . 10^-7 or 27/1000
Explanation:
We use the Coulomb Law
k = Coloumb Constant
q1 and q2 are the charges
d is the distance between the spheres
a fixed luminous point in the night sky which is a large, remote incandescent body like the sun. or a conventional or stylized representation of a star, typically one having five or more points.
<span>a.current varies throughout a parallel circuit.
Hope this helps!</span>
Explanation:
Equilibrium position in y direction:
W = Fb (Weight of the block is equal to buoyant force)
m*g = V*p*g
V under water = A*h
hence,
m = A*h*p
Using Newton 2nd Law
Hence, T time period
T = 2*pi*sqrt ( h / g )
Sure.
Can I use your answer to part-'a' ?
If the angular acceleration is actually 32 rev/min², than
after 1.2 min, it has reached the speed of
(32 rev/min²) x (1.2 min) = 38.4 rev/min .
Check:
If the initial speed is zero and the final speed is 38.4 rpm,
then the average speed during the acceleration period is
(1/2) (0 + 38.4) = 19.2 rpm average
At an average speed of 19.2 rpm for 1.2 min,
it covers
(19.2 rev/min) x (1.2 min) = 23.04 revs .
That's pretty close to the "23" in the question, so I think that
everything here is in order.
