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:
i think..its fraction that its have multiple fractions on it..if you minus the 397 000-355 it should be 381+ so i say if you get the 5 multiply it by 9!! so you will get it!
Explanation:
HOPE IT HELPS!!
Density <em>ρ</em> is mass <em>m</em> per unit volume <em>v</em>, or
<em>ρ</em> = <em>m</em> / <em>v</em>
Solving for <em>v</em> gives
<em>v</em> = <em>m</em> / <em>ρ</em>
So the given object has a volume of
<em>v</em> = (130 g) / (65 g/cm³) = 2 cm³
Power = (voltage) x (current)
1 watt = (1 volt) x (1ampere)
= 1 "volt-amp" or 1 "amp-volt" .
Answer:0.45ohms
Explanation:
Let R be there equivalent resistance
1/R=1/r+1/r+1/r
1/R=1/5+1/1+1/1
1/R=1/5+2
1/R=(1+10)/5
1/R=11/5
Cross multiplying we get
11R=5
Divide both sides by 11
11R ➗ 11=5 ➗ 11
R=0.45ohms
