Answer:
R = V / I
, R = V² / P, R = P / I²
Explanation:
For this exercise let's use ohm's law
V = I R
R = V / I
Electric power is defined by
P = V I
ohm's law
I = V / R
we substitute
P = V (V / R)
P = V² / R
R = V² / P
the third way of calculation
P = (i R) I
P = R I²
R = P / I²
Answer:
Writing an excellent problem statement will not help guide you through the rest of the process and steer you towards the BEST solution.
False
Explanation:
An excellent problem statement sets the overall tone for the rest of the engineering process, whether it be at the analysis, design, or implementation stages. This is why a problem statement must be focused, clear, and specific. An excellent problem statement contains the problem definition, method for solving the problem (the claim proposed), purpose, statement of objectives, and scope. For an excellent problem statement to be effective, it must also show the gap that is to be closed to achieve the intended objective.
Answer:
Q = 62 ( since we are instructed not to include the units in the answer)
Explanation:
Given that:
Q = ???
Now the gas expands at constant pressure until its volume doubles
i.e if 
Using Charles Law; since pressure is constant
mass of He =number of moles of He × molecular weight of He
mass of He = 3 kg × 4
mass of He = 12 kg
mass of Ar =number of moles of Ar × molecular weight of Ar
mass of He = 7 kg × 40
mass of He = 280 kg
Now; the amount of Heat Q transferred = 
From gas table
∴ Q = 
Q = 
Q = 62 MJ
Q = 62 ( since we are instructed not to include the units in the answer)
Answer:
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Explanation:
i have no idea what the question is but do u 4 real need help?
Answer:
a) 53 MPa, 14.87 degree
b) 60.5 MPa
Average shear = -7.5 MPa
Explanation:
Given
A = 45
B = -60
C = 30
a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)
P1 = 53 MPa
Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)
P1 = -68 MPa
Tan 2a = C/{(A-B)/2}
Tan 2a = 30/(45+60)/2
a = 14.87 degree
Principal stress
p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa
b) Shear stress in plane
Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa
Average = (45-(-60))/2 = -7.5 MPa
