Answer:
Vx = 6.242 x 10raised to power 15
Vy = -6.242 x 10raised to power 15
Explanation:
from E = IVt
but V = IR from ohm's law and Q = It from faraday's first law
I = Q/t
E = Q/t x V x t = QV
hence, E =QV
V = E/Q
Answer:
vx=6.242PV
vy=-6.242PV
Explanation:
Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the other terminal is the negative reference). The terminal on the right is the positive reference for a voltage called vy (the other terminal being the negative reference). If it takes 1 mJ of energy to push a single electron into the left terminal, determine the voltages vx and vy.
Here 1mj of energy is needed to an electron into the other terminal
voltage of an electron
Recall that E=IVt(product of power and time
but Q=quantity of charge is it (measured in coulombs)
E/Q=V
Ve=[tex]\frac{-1*10^{-3} }{-1.602*10^{-19} }[/tex]
0.6242*10^16
6.242*10^15
6.242PV(P=Peta=10^15)
voltage ve=6.242PV
Ve is used as Vy
The voltage vf is
Vf=[tex]\frac{1*10^{-3} }{-1.602*10^{-19} }[/tex]
Vf=-6.242*10^15
vf=-6.242PV
therefore, vf is designated as vy
the voltages are of the same magnitude but in different direction
