The detection of physical energy emitted or reflected by physical objects is called
1 answer:
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Answer:
( a ) The specific volume by ideal gas equation = 0.02632
% Error = 20.75 %
(b) The value of specific volume From the generalized compressibility chart = 0.0142
% Error = - 34.85 %
Explanation:
Pressure = 1 M pa
Temperature = 50 °c = 323 K
Gas constant ( R ) for refrigerant = 81.49
(a). From ideal gas equation P V = m R T ---------- (1)
⇒ =
⇒ Here = Specific volume = v
⇒ v =
Put all the values in the above formula we get
⇒ v = ×81.49
⇒ v = 0.02632
This is the specific volume by ideal gas equation.
Actual value = 0.021796
Error = 0.02632 - 0.021796 = 0.004524
% Error = × 100
% Error = 20.75 %
(b). From the generalized compressibility chart the value of specific volume
= v = 0.0142
The actual value = 0.021796
Error = 0.0142 - 0.021796 =
% Error = × 100
% Error = - 34.85 %
Answer:
= 85.89 ° C
Explanation:
The linear thermal expansion process is given by
ΔL = L α ΔT
For the three-dimensional case, the expression takes the form
ΔV = V β ΔT
Let's apply this equation to our case
ΔV / V = -0.507% = -0.507 10-2
ΔT = (ΔV / V) 1 /β
ΔT = -0.507 10⁻² 1 / 1.15 10⁻³
ΔT = -4.409
–T₀ = 4,409
= T₀ - 4,409
= 90.3-4409
= 85.89 ° C
Answer:
a) E=228391.8 N/C
b) E=-59345.91N/C
Explanation:
You can use Gauss law to find the net electric field produced by both line of charges.
Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.
The net electric field at point r will be:
a) for y=0.200m, r1=0.200m and r2=0.200m:
b) for y=0.600m, r1=0.600m, r2=0.200m: