Answer:
Modulus of resilience will be ![3216942.308j/m^3](https://tex.z-dn.net/?f=3216942.308j%2Fm%5E3)
Explanation:
We have given yield strength ![\sigma _y=818MPa](https://tex.z-dn.net/?f=%5Csigma%20_y%3D818MPa)
Elastic modulus E = 104 GPa
We have to find the modulus
Modulus of resilience is given by
Modulus of resilience
, here
is yield strength and E is elastic modulus
Modulus of resilience
Answer:
When the expenditure increased, then the consumer's expenditure is increased to Rs.150 and when the price falls of the good it becomes Rs.5. Then, Good X will be Rs.10.
last question:If the Good X falls by 20%, then, it will be Rs.2, and according to his demand 100 units will be equal to Rs.200.
cause if one unit=rs.2, then 100units=100×2=200.
Solution :
Given :
The number of blows is given as :
0 - 6 inch = 4 blows
6 - 12 inch = 6 blows
12 - 18 inch = 6 blows
The vertical effective stress ![$=1500 \ lb/ft^2$](https://tex.z-dn.net/?f=%24%3D1500%20%5C%20lb%2Fft%5E2%24)
![$= 71.82 \ kN/m^2$](https://tex.z-dn.net/?f=%24%3D%2071.82%20%5C%20kN%2Fm%5E2%24)
![$ \sim 72 \ kN/m^2 $](https://tex.z-dn.net/?f=%24%20%5Csim%2072%20%5C%20kN%2Fm%5E2%20%24)
Now,
![$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$](https://tex.z-dn.net/?f=%24N_1%3DN_0%20%5Cleft%28%5Cfrac%7B350%7D%7B%5Cbar%7B%5Csigma%7D%2B70%7D%20%5Cright%29%24)
corrected N - value of overburden
effective stress at level of test
0 - 6 inch, ![$N_1=4 \left(\frac{350}{72+70} \right)$](https://tex.z-dn.net/?f=%24N_1%3D4%20%5Cleft%28%5Cfrac%7B350%7D%7B72%2B70%7D%20%5Cright%29%24)
= 9.86
6 - 12 inch, ![$N_1=6 \left(\frac{350}{72+70} \right) $](https://tex.z-dn.net/?f=%24N_1%3D6%20%5Cleft%28%5Cfrac%7B350%7D%7B72%2B70%7D%20%5Cright%29%20%24)
= 14.8
12 - 18 inch, ![$N_1=6 \left(\frac{350}{72+70} \right) $](https://tex.z-dn.net/?f=%24N_1%3D6%20%5Cleft%28%5Cfrac%7B350%7D%7B72%2B70%7D%20%5Cright%29%20%24)
= 14.8
![$N_{avg}=\frac{9.86+14.8+14.8}{3}$](https://tex.z-dn.net/?f=%24N_%7Bavg%7D%3D%5Cfrac%7B9.86%2B14.8%2B14.8%7D%7B3%7D%24)
= 13.14
= 13
Answer:
O is truse is the best answer hhahahha
Explanation:
Answer:
Hello your question has some missing information below are the missing information
The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump
answer : 2.49
Explanation:
For vapor-compression refrigeration cycle
P1 = P4 ; P1 = 140 kPa
P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa
<u>From pressure table of R 134a refrigerant</u>
h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg
h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )
= 296.8kJ/kg
h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg
also h4 = 95.47 kJ/kg
To determine the coefficient of performance
Cop = ( h1 - h4 ) / ( h2 - h1 )
∴ Cop = 2.49