Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the o
ther 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.
2 answers:
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=
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=
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
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Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 /
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
=
given that = 480 mm
we substitute
=
×
= 501.998 mm
Now we find the elongation;
Elongation =
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Answer:
# -*- coding: utf-8 -*-
# Get N from the command line
import sys
N = int(sys.argv[1])
if N > 0: #N is positive
positve = list(range(N,0,-1))
print(positve)
elif N < 0: #N is negative
negative = list(range(N,0,1))
print(negative)
else:
print("Invalid in input")
Explanation:
First, you need to identify if the number entered is positive, negative or none of them, for that we use one if, one elif and one else statement:
- If the number entered (N) is greater than zero (is positive) we print a list in the range N to 0 in steps of minus one, the zero is not printed because the function range by default omits the last value
- If the previous statement was False and the number entered is smaller than zero (is negative) we print a list in the range N to 0 in steps of one, the zero is not printed because the function range by default omits the last value
- Finally, if the two previous events were false print invalid input
