Show Clear and detailed calculation for all problems. DO NOT put solutions on th
ID: 1446703 • Letter: S
Question
Show Clear and detailed calculation for all problems. DO NOT put solutions on the test paper. IT WILL NOT BE GRADED Data: V= IR; P = VI; P = l^2R ; R = rhoL/A EPE = q delta V E = kq/r^2 v = kq/r A massive spool of copper 100 km has a diameter of 0.5 cm. As copper has great value it is stored in a secured vault. However, a gang of thieves gets into the valut and hacks a considerable length. Tests reveal that the remaining cable has a resistance of 38 Ohm. [rho = 1.73 x 10^-8 Ohm-meter for copper] Determine the length of cable stolen by the gang. Calculate the ratio of the length of cable remaining to the origial length. Two charges q1 = -15 muC and q2 = +20 muC arc 4.0 m apart as shown in the diagram below. Compute the work (electrostatic potential energy) that must be done to bring a third charge q3 = + 8.0 muC from infinity to the point P, 3.0 m directly above q1.Explanation / Answer
L1 = origiinal length = 100 km
d = diameter = 0.5 cm
r = radius = 0.5 *10-2 / 2 = 2.5*10-3m
R2 = remaining length resiatance = 38 ohm
R1 = original length resistance = p* L1 / A ( p = resistivity , A = area)
R1 = 1.73*10-8*100*103 / 3.14*(2,5*10-3)2
R1 = 88.15 ohm
R2 = p*L2 / A ( L2 = length of cable stolen)
R1 / R2 = (p*L1 / A ) / ( p*L2 / A)
88.15 / 38 = 100*103 ?L2
L2 = 43.1 km
ii) remaining length = 100 - 43.1 = 56.9 km
remaining length / original length = 56.9 km / 100 km
= 0.569