The potential difference between the two points if the work done on the charge is +3.4×10^7 joules would be A. 10volts.
Complete question is;
A copper wire has a diameter of 4.00 × 10^(-2) inches and is originally 10.0 ft long. What is the greatest load that can be supported by this wire without exceeding its elastic limit? Use the value of 2.30 × 10⁴ lb/in² for the elastic limit of copper.
Answer:
F_max = 28.9 lbf
Explanation:
Elastic limit is simply the maximum amount of stress that can be applied to the wire before it permanently deform.
Thus;
Elastic limit = Max stress
Formula for max stress is;
Max stress = F_max/A
Thus;
Elastic limit = F_max/A
F_max is maximum load
A is area = πr²
We have diameter; d = 4 × 10^(-2) inches = 0.04 in
Radius; r = d/2 = 0.04/2 = 0.02
Plugging in the relevant values into the elastic limit equation, we have;
2.30 × 10⁴ = F_max/(π × 0.02²)
F_max = 2.30 × 10⁴ × (π × 0.02²)
F_max = 28.9 lbf
-- Sound energy. That's how you hear it explode.
-- Light energy. That's how you see it in the sky.
-- Heat energy. That's why you get burned if sparks fall on you.
-- Potential energy, as all the pieces fall to the ground.
These were all stored in it as chemical energy before it exploded.
Explanation:
object made of rubber, such as the ballons are elecrical insulators, meaning that they resist electric charges flowing through them..
when the ballon has been rubbed enough times to gain a sufficient negative charge, it will be attracted to the wall
Answer:
b it is b part answer i think so
