A 17.5-kg block is dragged over a rough, horizontal surface by a 75-N force acting at 21° above the horizontal. The block is dis
placed 5.7 m, and the coefficient of kinetic friction is 0.150.
Requried:
a. Find the work done on the block by the 75-N force.
b. Find the work done on the block by the normal force.
c. Find the work done on the block by the gravitational force.
d. What is the increase in internal energy of the block-surface system due to friction?
e. Find the total change in the block's kinetic energy.
Requried:
a. Find the work done on the block by the 75-N force.
b. Find the work done on the block by the normal force.
c. Find the work done on the block by the gravitational force.
d. What is the increase in internal energy of the block-surface system due to friction?
e. Find the total change in the block's kinetic energy.
1 answer:
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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:
f = 931.1 Hz
Explanation:
Given,
Mass of the wire, m = 0.325 g
Length of the stretch, L = 57.7 cm = 0.577 m
Tension in the wire, T = 650 N
Frequency for the first harmonic = ?
we know,
μ is the mass per unit length
μ = 0.325 x 10⁻³/ 0.577
μ = 0.563 x 10⁻³ Kg/m
now,
v = 1074.49 m/s
The wire is fixed at both ends. Nodes occur at fixed ends.
For First harmonic when there is a node at each end and the longest possible wavelength will have condition
λ=2 L
λ=2 x 0.577 = 1.154 m
we now,
v = f λ
f = 931.1 Hz
The frequency for first harmonic is equal to f = 931.1 Hz