Answer:
A) 60%
B) p2 = 1237.2 kPa
v2 = 0.348 m^3
C) w1-2 = w3-4 = 1615.5 kJ
Q2-3 = 60 kJ
Explanation:
A) calculate thermal efficiency
Л = 1 -
where Tl = 300 k
Th = 750 k
hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%
B) calculate the pressure and volume at the beginning of the isothermal expansion
calculate pressure ( P2 ) :
= P3v3 = mRT3 ----- (1)
v3 = 0.4m , mR = 2* 0.287, T3 = 750
hence P3 = 1076.25
next equation to determine P2
Qex = p3v3 ln( p2/p3 )
60 = 1076.25 * 0.4 ln(p2/p3)
hence ; P2 = 1237.2 kpa
calculate volume ( V2 )
p2v2 = p3v3
v2 = p3v3 / p2
= (1076.25 * 0.4 ) / 1237.2
= 0.348 m^3
C) calculate the work and heat transfer for each four processes
work :
W1-2 = mCv( T2 - T1 )
= 2*0.718 ( 750 - 300 ) = 1615.5 kJ
W3-4 = 1615.5 kJ
heat transfer
Q2-3 = W2-3 = 60KJ
Q3-4 = 0
D ) sketch of the cycle on p-V coordinates
attached below
Answer:
700 mL or 0.0007 m³
Explanation:
P₁ = Initial pressure = 2 atm
V₁ = Initial volume = 350 mL
P₂ = Final pressure = 1 atm
V₂ = Final volume
Here the temperature remains constant. So, Boyle's law can be applied here.
P₁V₁ = P₂V₂

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³
1) Blood flow: increases during warming improving muscle and joint elasticity. This decreases the possibility of having an injury.
2) Body temperature: This causes the cellular metabolism to increase. It also causes vasodilatation that allows a greater supply of oxygen and nutrients.
The change in speed of this object is 3m/s
According to Newton's second law;
F = ma
F = mv/t
Given the following parameters
Force F = 8.0N
mass m = 16kg
time t = 4.0s
Required
speed v
Substitute the given parameters into the formula
v = Ft/m
v = 8 * 6/16
v = 48/16
v = 3m/s
Hence the change in speed of this object is 3m/s
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