Answer:
1 joule = 0.737 foot-pound
Joule is the unit of work.
1 J = 1 N·m
In SI units
1 J = 1 kg· m/s²
0.737 foot-pound is the amount of work to raise 0.737 pounds one foot or raising one pound to 0.737 ft.
Answer:
Explanation:
For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:
which can also be rewritten as
In our case, we have:
is the initial pressure
is the initial volume
is the final pressure
Solving for V2, we find the final volume:
Answer:
422.36 N
Explanation:
given,
time of rotation = 4.30 s
T = 4.30 s
Assuming the diameter of the ring equal to 16 m
radius, R = 8 m
v = 11.69 m/s
now, Force does the ring push on her at the top
N = 422.36 N
The force exerted by the ring to push her is equal to 422.36 N.
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
The answer is to increase energy. Hope this helps!
