Answer:
Work done = 125π J
Explanation:
Given:
P = P_i * ( 1 - (x/d)^2 / 25)
d = 5.0 cm
x = 5 * d cm = 25d
Pi = 12 bar
Work done = integral ( F . dx )
F (x) = P(x) * A
F (x) = (πd^2 / 4) * P_i * (1 - (x/d)^2 / 25)
Work done = integral ((πd^2 / 4) * P_i * (1 - (x/d)^2 / 25) ) . dx
For Limits 0 < x < 5d
Work done = (πd^2 / 4) * P_i integral ( (1 - (x/d)^2) / 25)) . dx
Integrate the function wrt x
Work done = (πd^2 / 4) * P_i * ( x - d*(x/d)^3 / 75 )
Evaluate Limits 0 < x < 5d :
Work done = (πd^2 / 4) * P_i * (5d - 5d / 3)
Work done = (πd^2 / 4) * P_i * (10*d / 3)
Work done = (5 π / 6)d^3 * P_i
Input the values:
Work done = (5 π / 6)(0.05)^3 * (1.2*10^6)
Work done = 125π J
Answer:
Correct option a) True.
Explanation:
It is true since the Vickers hardness value refers to the force applied in a 136 ° diamond tip penetrator divided by the surface of the groove produced in the material, the lower the impression made on this greater the value will be end of the Vickers measurement and greater its hardness.
The equation to determine the Vickers hardness value will be:
Hv= ((1.854 × P)/(d²)) (kg/mm²)
Therefore a value of 220 Vickers refers to a harder material than another value of 180 Vickers.
Answer: P = 0.416 kW
Explanation:
taken a step by step process to solving this problem.
we have that from the question;
the amount of heat rejected Qn = 4800 kJ/h
the cooling effect is Ql = 3300 kJ/h
Applying the first law of thermodynamics for this system gives us
Шnet = Qn -Ql
Шnet = 4800 - 3300 = 1500 kJ/h
Next we would calculate the coefficient of performance of the refrigerator;
COPr = Desired Effect / work output = Ql / Шnet = 3300/1500 = 2.2
COPr = 2.2
The Power as required gives;
P = Qn - Ql = 4800 - 3300 = 1500 kJ/h = 0.416
P = 0.416 kW
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