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:
The minimum value of wall thickness t=3.63 mm.
Explanation:
Given:
D=200 mm
P=4 MPa
t= Wall thickness
maximum shear stress=27.5 MPa
We know that
hoop stress 
Longitudinal stress
So maximum shear tress in plane

Now by putting the value

So t=3.36 mm
The minimum value of wall thickness t=3.63 mm.
The modulus of elasticity is 28.6 X 10³ ksi
<u>Explanation:</u>
Given -
Length, l = 5in
Force, P = 8000lb
Area, A = 0.7in²
δ = 0.002in
Modulus of elasticity, E = ?
We know,
Modulus of elasticity, E = σ / ε
Where,
σ is normal stress
ε is normal strain
Normal stress can be calculated as:
σ = P/A
Where,
P is the force applied
A is the area of cross-section
By plugging in the values, we get
σ = 
σ = 11.43ksi
To calculate the normal strain we use the formula,
ε = δ / L
By plugging in the values we get,
ε = 
ε = 0.0004 in/in
Therefore, modulus of elasticity would be:

Thus, modulus of elasticity is 28.6 X 10³ ksi
Answer: The exit temperature of the gas in deg C is
.
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa,
We know that for an ideal gas the mass flow rate will be calculated as follows.
or, m =
=
= 10 kg/s
Now, according to the steady flow energy equation:
= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
=
Therefore, we can conclude that the exit temperature of the gas in deg C is
.
Answer:
not sure if this helps but