A machine has an efficiency of 15%. If the energy input is 300 joules, how much useful energy is generated?(1 point).
1 answer:
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Answer:
hello your question is incomplete attached below is the complete question
answer : Drag force = 1.3 Ib
Explanation:
we have to represent the dimensions of the drag force in terms of FLT
i.e : D = f( <em>d,v,p,u </em>) represented in terms of FLT
D = F , V = LT^-1, d = L, p = FL^-4 T^2
u = FL^-2 T, Number of independent terms = 5
attached below is the detailed solution
Answer:C 0.12 V
Explanation:
Given
Concentration of
Concentration of
Standard Potential for Ni and Fe are and
Answer:
a) 358.8K
b) 181.1 kJ/kg.K
c) 0.0068 kJ/kg.K
Explanation:
Given:
P1 = 100kPa
P2= 800kPa
T1 = 22°C = 22+273 = 295K
q_out = 120 kJ/kg
∆S_air = 0.40 kJ/kg.k
T2 =??
a) Using the formula for change in entropy of air, we have:
∆S_air =
Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K
Solving, we have:
[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]
Solving for T2 we have:
Taking the exponential on the equation (both sides), we have:
[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]
b) Work input to compressor:
= 184.1 kJ/kg
c) Entropy genered during this process, we use the expression;
Egen = ∆Eair + ∆Es
Where; Egen = generated entropy
∆Eair = Entropy change of air in compressor
∆Es = Entropy change in surrounding.
We need to first find ∆Es, since it is unknown.
Therefore ∆Es =
∆Es = 0.4068kJ/kg.k
Hence, entropy generated, Egen will be calculated as:
= -0.40 kJ/kg.K + 0.40608kJ/kg.K
= 0.0068kJ/kg.k