Answer:
Explanation:
The amplitude of he combined wave is:
A, is the amplitude from the identical harmonic waves
B, is the amplitude of the resultant wave
θ, is the phase, between the waves
The amplitude of the combined wave must be 0.6A:
Because of how it's worded the answer would most likely be number four
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
Weight=mg
g=GM/r^2
g on venus is 0.80w as radius is kept constant
m of object is kept constant
w α g
w(venus( is 0.8w
Work done = force * distance moved (in direction of the force)
force= mass* acceleration
force=58.1N
58.1*(5.8*10^4)
=3,369,800 J
