Could anyone answer this, please? It's about solid mechanics. I will give you 100 points!!! It's due at midnight.
Answers
Answer:
sorry i don't know
Explanation:
Related Questions
In a ground-water basin of 12 square miles, there are two aquifers: an upper unconfined aquifer 500 ft in thickness and a lower confined aquifer with an available hydraulic head drop of 150 ft. Hydraulic tests have determined that the specific yield of the upper unit is 0.12 and the storativity of the lower unit is 4x10-4. What is the amount of recoverable ground water in the basin
Answers
Answer:
0.1365 m^3
Explanation:
thickness of upper aquifer = 500 ft
lower aquifer head drop = 150 ft
area of ground water basin = 12 m^2
specific yield of upper unit = 0.12
Storativity of lower unit = 4 * 10^-4
determine the amount of recoverable ground water
first step : calculate volume of unconfined aquifer
= 12 * 500/5280 = 1.1364 miles^3
The recoverable volume of water from unconfined aquifer
= 1.1364 * 0.12 = 0.1364 miles^3
next : calculate volume of confined aquifer
= 12 * 150/5250 = 0.341 miles^3
The recoverable volume of water from confined aquifer
= 0.341 * ( 4 * 10^-4 ) = 1.364 * 10^-4 miles^3
Hence the amount of recoverable ground water in the basin
= ∑ recoverable ground water from both aquifer
= 0.1365 m^3
Air at 308C, 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate of 3 kg/min and mixes with a saturated moist air stream entering at 58C, 1 bar with a mass flow rate of 5 kg/min. A single mixed stream exits at 1 bar. Determine (a) the relative humidity and temperature, in 8C, of the exiting stream. (b) the rate of exergy destruction, in kW, for T0 5 208C. Neglect kinetic and potential energy effects.
Answers
Answer: the question of whether or is that a new place for a person to come in the morning and then the day that we have a good day at school or to come home with us to go to the church or we could meet up at my place in about a week or so to get the rest of the kids and I can go out to the school to go to the gym to go to the doctor to pick them out or not I have a good time to come over to get the stuff out of the car so I’m going out of the house to go to the store to pick it out I don’t have any money for that
Explanation:
forty gal/min of a hydrocarbon fuel having a spesific gravity of 0.91 flow into a tank truck with load limit of 40,000 lb of fuel. How long will it takee to fill the tank in the truck?
Answers
Answer: 131.75minutes
Explanation:
First if all, we've to find the density of liquid which will be:
= Specific gravity × Density to pure water
= 0.91 × 8.34lb/gallon
= 7.59lb/gallon
Then, the volume that's required to fill the tank will be:
= Load limit/Density of fluid
= 40000/7.59
= 5270.1gallon
Now, the time taken will be:
= V/F
= 5270.1/40
= 131.75min
It'll take 131.75 minutes to fill the tank in the truck.
Q2 [45 marks] Consider Ibra region where the installed solar panels cost on average 2 OMR /W.
[10 marks] What is the cost to install a 5kW PV system for a residence?
[10 marks] If the solar irradiance in Ibra is on average 800W/m2 and the installed panels have efficiency of 18%. How many panels are required if the panel’s area is 2m2?
[15 marks] Assume Ibra has an average of 10 day-hours, dusty environment which causes the efficiency of the solar system to drop by 10% on average, and 30 cloudy days/year which cause the efficiency of the solar panels drops by 50%. If electrical power cost per kWh is 0.05 OMR determine the break-even time for the 5kW PV system.
[10 marks] If the system to be off-grid, what would be the backup time if three 12-V batteries were selected each with a capacity of 200Ah. Assume that you can discharge the batteries up to 80% of their capacities.
Answers
Answer:
so hard it is
Explanation:
I don't know about this
please mark as brainleast
byýyy
A cylindrical rod of copper (E = 110 GPa) having a yield strength of 240 MPa is to be subjected
to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an
elongation of 0.50 mm?
Answers
Answer:
"7.654 mm" is the correct solution.
Explanation:
According to the question,
[tex]E=110\times 10^3 \ N/mm^2[/tex][tex]\sigma_y = 240 \ mPa[/tex][tex]P = 6660 \ N[/tex][tex]L = 380 \ mm[/tex][tex]\delta = 0.5 \ mm[/tex]Now,
As we know,
The Elongation,
⇒ [tex]E=\frac{\sigma}{e}[/tex]
[tex]=\frac{\frac{P}{A} }{\frac{\delta}{L} }[/tex]
or,
⇒ [tex]\delta=\frac{PL}{AE}[/tex]
By substituting the values, we get
[tex]0.5=\frac{6660\times 380}{(\frac{\pi}{4}D^2)(110\times 10^3)}[/tex]
then,
⇒ [tex]D^2=58.587[/tex]
[tex]D=\sqrt{58.587}[/tex]
[tex]=7.654 \ mm[/tex]