Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg
Answer:
Explanation:
From the question we are told that:
Mass 
Angle 
Coefficient of static friction 
Generally, the equation for Newtons second Law is mathematically given by
For
for
Where
Therefore
Answer
Applying Wein's displacement
1) for sun T = 5800 K
2) for tungsten T = 2500 K
3) for heated metal T = 1500 K
4) for human skin T = 305 K
5) for cryogenically cooled metal T = 60 K
range of different spectrum
UV ----0.01-0.4
visible----0.4-0.7
infrared------0.7-100
for sun T = 5800
λ 0.01 0.4 0.7 100
λT 58 2320 4060 5.8 x 10⁵
F 0 0.125 0.491 1
fractions
for UV = 0.125
for visible = 0.441-0.125 = 0.366
for infrared = 1 -0.491 = 0.509
Answer:
It is most likely option A B and C
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
