If an object gains an additional 5 m/s of speed every second, the object is?
2 answers:
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Answer:
The Entropy generated by the steam = 2.821 kJ/K
Explanation:
Total volume of container = 5m³
Heat transfer does not exist between system and surrounding, dQ = 0
At the first chamber, temperature of water at saturated liquid is 300°C
From the steam table:
Specific enthalpy of saturated liquid at 300°C ,
Specific internal energy of saturated liquid at 300°C, 1332.7 kJ/kg
For closed system, the first law of thermodynamics state that:
dQ = dw + dU..................(1)
work done for free expansion, dw =0
0 = 0 + dU
dU = 0 , i.e. U₁ = U₂
At the second chamber,
The final pressure, P₂ = 50 kPa
From the steam table, at P₂ = 50 kPa, = 340.49 kJ/kg
Let the dryness fraction at the second chamber = x
Since U₁ = U₂
1332.7 = 340.49 + x2140.7
Dryness fraction, x = 0.463
From steam table, the specific volume is,
V₂ = 5 m³
1.5 = 5/m₂
m₂ = 3.33 kg
At 300°C
From the steam table,
Therefore the entropy generated will be :
Entropy = mass* (S₂ - S₁)
Entropy = 3.33* (4.102 - 3.2548)
Entropy = 2.821 kJ/K
Answer:
(a) 89 m/s
(b) 11000 N
Explanation:
Note that answers are given to 2 significant figures which is what we have in the values in the question.
(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.
(b) The tension, , is given by
where is the speed,
is the tension and
is the mass per unit length.
Hence,
To determine , we need to know the mass of the cable. We use the density formula:
where is the mass and
is the volume.
If the length is denoted by , then
The density of steel = 8050 kg/m3
The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is