Answer:
$958.25
Explanation:
PV = FV(1 +r)^(-t) . . . . present value of an amount earning rate r compounded annually for t years
PV = $3000(1.08^-5) ≈ $2041.75
The equivalent present value is $2041.75.
__
This amount is less than the offered amount by ...
$3000 -2041.75 = $958.25
She should pay you $958.25 less in order to make the offers equivalent.
Answer:
![h = 1429.74\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h%20%3D%201429.74%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
Explanation:
The determination of any further properties requires the knowledge of two independent properties. (Temperature and specific volume in this case). The specific volumes for saturated liquid and vapor at 140 °C are, respectively:
![\nu_{f} = 0.001080\,\frac{m^{3}}{kg}](https://tex.z-dn.net/?f=%5Cnu_%7Bf%7D%20%3D%200.001080%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%7D)
![\nu_{g} = 0.50850\,\frac{m^{3}}{kg}](https://tex.z-dn.net/?f=%5Cnu_%7Bg%7D%20%3D%200.50850%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%7D)
Since
, it is a liquid-vapor mixture. The quality of the mixture is:
![x = \frac{\nu-\nu_{f}}{\nu_{g}-\nu_{f}}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B%5Cnu-%5Cnu_%7Bf%7D%7D%7B%5Cnu_%7Bg%7D-%5Cnu_%7Bf%7D%7D)
![x = \frac{0.2\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }{0.50850\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B0.2%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%7D%20-%200.001080%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%7D%20%7D%7B0.50850%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%7D%20-%200.001080%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%7D%20%7D)
![x = 0.392](https://tex.z-dn.net/?f=x%20%3D%200.392)
The specific enthalpies for saturated liquid and vapor at 140 °C are, respectively:
![h_{f} = 589.16\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h_%7Bf%7D%20%3D%20589.16%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
![h_{g} = 2733.5\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h_%7Bg%7D%20%3D%202733.5%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
The specific enthalpy is:
![h = h_{f}+x\cdot (h_{g}-h_{f})](https://tex.z-dn.net/?f=h%20%3D%20h_%7Bf%7D%2Bx%5Ccdot%20%28h_%7Bg%7D-h_%7Bf%7D%29)
![h = 589.16\,\frac{kJ}{kg}+0.392\cdot \left( 2733.5\,\frac{kJ}{kg} - 589.16\,\frac{kJ}{kg} \right)](https://tex.z-dn.net/?f=h%20%3D%20589.16%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D%2B0.392%5Ccdot%20%5Cleft%28%202733.5%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D%20-%20589.16%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D%20%5Cright%29)
![h = 1429.74\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h%20%3D%201429.74%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
Answer:
A) 2P/(¶r^2)
B) p/(¶r^2)
Explanation:
Force = P
Radius = r
Area on which force acts = ¶r^2
If both forces are applied, total stress will be
S = 2P/(¶r^2)
When only one of the force is applied
S = p/(¶r^2)
Answer: Temperature T = 218.399K.
Explanation:
Q= 129W/m2
σ = 5.67 EXP -8W/m2K4
Q= σ x t^4
t = (Q/σ)^0.25
t= 218.399k