Answer:
R = 715.4 N
L = 166.6 N
Explanation:
ASSUME the painter is standing right of center
Let L be the left rope tension
Let R be the right rope tension
Sum moments about the left end to zero. Assume CCW moment is positive
R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0
R = 715.4 N
Sum moments about the right end to zero
20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0
L = 166.6 N
We can verify by summing vertical forces
116.6 + 715.4 - (70 + 20)(9.8) ?=? 0
0 = 0 checks
If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.
Answer:
Im answering for free points sry
Explanation:
...
Answer:
The value is 
Explanation:
From the question we are told that
The radius of the inner conductor is 
The radius of the outer conductor is 
The potential at the outer conductor is 
Generally the capacitance per length of the capacitor like set up of the two conductors is
![C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}](https://tex.z-dn.net/?f=C%3D%20%5Cfrac%7B2%20%2A%20%5Cpi%20%2A%20%5Cepsilon_o%20%7D%7B%20ln%20%5B%5Cfrac%7Br_2%7D%7Br_1%7D%20%5D%7D)
Here
is the permitivity of free space with value 
=> ![C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}](https://tex.z-dn.net/?f=C%3D%20%5Cfrac%7B2%20%2A%20%203.142%20%20%2A%208.85%2A10%5E%7B-12%7D%20%20%7D%7B%20ln%20%5B%5Cfrac%7B0.003%7D%7B0.001%7D%20%5D%7D)
=> 
Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge
Generally the line charge density of the outer conductor is mathematically represented as

=> 
=> 
Generally the surface charge density is mathematically represented as
here 
=> 
=> 
The answer is B, because all energy is released at once in static electricity.
There’s a quizlet that mentions these questions, if you are having trouble. I’d suggest to give them a look.