Answer:
What's the purpose of tracks going in the red? Having tracks go into the red is surely redundant, I don't see any purpose in having tracks distort ... It just seems like a hang on from the old days of tape, it's something that people who ... be in daws and I'm trying to assemble an alternative I understand the current mixing system. The Dow Jones Industrial Average (DJIA) is a stock index of 30 blue-chip industrial ... Today, the DJIA is a benchmark that tracks American stocks that are ... To calculate the DJIA, the current prices of the 30 stocks that make up the ... the longevity of the Dow serves this purpose better than all other indices.
Explanation:
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
Answer:
hello some parts of your question is missing attached below is the missing part ( the required fig and table )
answer : The solar collector surface area = 7133 m^2
Explanation:
Given data :
Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2
percentage of solar power absorbed by refrigerant = 60%
Determine the solar collector surface area
The solar collector surface area = 7133 m^2
attached below is a detailed solution of the problem
Answer:
The spring is compressed by 0.275 meters.
Explanation:
For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of weight of the piston and the force the spring exerts on the piston
Mathematically we can write
![Force_{pressure}=Force_{spring}+Weight_{piston}](https://tex.z-dn.net/?f=Force_%7Bpressure%7D%3DForce_%7Bspring%7D%2BWeight_%7Bpiston%7D)
we know that
![Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons](https://tex.z-dn.net/?f=Force_%7Bpressure%7D%3DPressure%5Ctimes%20Area%3D300%5Ctimes%2010%5E%7B3%7D%5Ctimes%20%5Cfrac%7B%5Cpi%20%5Ctimes%200.1%5E2%7D%7B4%7D%3D750%5Cpi%20Newtons)
![Weight_{piston}=mass\times g=100\times 9.81=981Newtons](https://tex.z-dn.net/?f=Weight_%7Bpiston%7D%3Dmass%5Ctimes%20g%3D100%5Ctimes%209.81%3D981Newtons)
Now the force exerted by an spring compressed by a distance 'x' is given by ![Force_{spring}=k\cdot x=5\times 10^{3}\times x](https://tex.z-dn.net/?f=Force_%7Bspring%7D%3Dk%5Ccdot%20x%3D5%5Ctimes%2010%5E%7B3%7D%5Ctimes%20x)
Using the above quatities in the above relation we get
![5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters](https://tex.z-dn.net/?f=5%5Ctimes%2010%5E%7B3%7D%5Ctimes%20x%2B981%3D750%5Cpi%20%5C%5C%5C%5C%5Ctherefore%20x%3D%5Cfrac%7B750%5Cpi%20-981%7D%7B5%5Ctimes%2010%5E%7B3%7D%7D%3D0.275meters)