Answer:
the size of the shadow will be smaller due to smaller hands
<span>D. density is your answer</span>
Answer:
84.82N/C.
Explanation:
The x-components of the electric field cancel; therefore, we only care about the y-components.
The y-component of the differential electric field at the center is
.
Now, let us call 
the charge per unit length, then we know that
;
therefore,
Integrating
![$E = \frac{k \lambda }{R}*[-cos(\pi )+cos(0) ]$](https://tex.z-dn.net/?f=%24E%20%3D%20%5Cfrac%7Bk%20%5Clambda%20%20%20%7D%7BR%7D%2A%5B-cos%28%5Cpi%20%29%2Bcos%280%29%20%5D%24)
Now, we know that
and the radius of the semicircle is
therefore,
5kg
50cm
500in
Hope this helped good luck to you
The first thing you should know for this case is that work is defined as the product of force by the distance traveled in the direction of force.
We have then:
W = Fd
The distance varies, so we must integrate:
from 0 to 20:
W = ∫F (x) dx
W = ∫32xdx
W = 32∫xdx
W = 32 (x ^ 2/2) = (16) (20 ^ 2) = 6400 ft * lbs
answer:
6400 ft * lbs is work done pulling the rope up 20 ft
