Since the law of gravitation is an inverse square law if you
quadruple the radius the f will drop by a factor of 16 SO the object would
weigh 200/16 = 12.5N
In other words, as the distance, or radius, quadruples the
weight becomes 1/16 of the original weight. Just plug in 4 for r and when you
square it you get 16. The numerator is 1 so that is how the weight becomes
1/16.
This problem refers to a parallel plate capacitor. There is
an electric field between the two plates. The working equation to be used is
the Gauss’s Law which is
Electric field = Surface charge density / ε0
The answer is -2.52 μC/m2.
The weight on Earth would be:
77.3 * 9.8
= 757.54 Newtons
1 Newton = 0.2248 lbf
757.54 * 0.2248
= 170.29 lbf
On the asteroid:
77.3 * 0.08
= 6.18 Newtons
Applying the same conversion:
77.3 * 0.2240
= 17.31 lbf
Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 =
ω2 = 216.47