The high spark voltage supplied to the spark plugs of an automobile is increased from the battery voltage by the principle of .
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Answer:
Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.
Explanation:
<u>Objects at terminal velocity</u>, only under the influence of gravity, have maximized their speed and <u>have an acceleration of zero</u>. Thus, neither object is accelerating.
Recall Newton's second law:
Since acceleration for each object is zero, the sum of the force acting on each of those objects must also be zero.
Since the only forces acting on the objects are gravity and the force of air friction, in order to zero out, <u>the force of air friction must be equal in magnitude and opposite in direction to the force of gravity</u>.
Recall that near the surface of the earth, , so <u>the Force of Gravity acting on an object is directly proportional to the object's mass</u>. <em>(A similar argument could be made even if this were not taking place on the surface of the earth, so long as the objects were the same distance from the object providing gravitational influence).</em>
If the masses of the objects are different, <u>the object with the greater mass will experience</u> a larger force of gravity, and hence <u>a larger force of air friction</u> at terminal velocity.
Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.
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