Answer:
The elevation at the high point of the road is 12186.5 in ft.
Explanation:
The automobile weight is 2500 lbf.
The automobile increases its gravitational potential energy in 
. It means the mobile has increased its elevation.
The initial elevation is of 5183 ft.
The first step is to convert Btu of potential energy to adequate units to work with data previously presented.
British Thermal Unit - 
Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:
Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.
is the final elevation and 
is the initial elevation.
Replacing W in the Ep equation
Finally, the next step is to replace the variables of the problem.
The elevation at the high point of the road is 12186.5 in ft.
Answer:
Tesla CEO Elon Musk tweeted that the company's stock was too high, and it immediately dropped in value. The tweet may have violated a deal Musk made with the SEC about his tweets and Tesla. Musk also tweeted patriotic lyrics, said his girlfriend Grimes is mad at him, and noted that their child is due on Monday
Answer:
Responsibility
Explanation:
By stamping the drawings that he was looking over, Jack Gillum conveys the fact that he is accepting responsibility for this work. The purpose of Gillum's stamp is to explain that such work has been under engineering review, and that it has fulfilled all the requirements that he watches our for. By putting his stamp in this work, Gillum accepts responsibility in case an error or a discrepancy is found in the drawings.
Answer: I would love to learn this
Explanation:
Answer:
η=0.19=19% for p=14.7psi
η=0.3=30% for p=1psi
Explanation:
enthalpy before the turbine, state: superheated steam
h1(p=200psi,t=500F)=2951.9KJ/kg
s1=6.8kJ/kgK
Entalpy after the turbine
h2(p=14.7psia, s=6.8)=2469KJ/Kg
Entalpy before the boiler
h3=(p=14.7psia,x=0)=419KJ/Kg
Learn to pronounce
the efficiency for a simple rankine cycle is
η=
η=(2951.9KJ/kg-2469KJ/Kg)/(2951.9KJ/kg-419KJ/Kg)
η=0.19=19%
second part
h2(p=1psia, s=6.8)=2110
h3(p=1psia, x=0)=162.1
η=(2951.9KJ/kg-2110KJ/Kg)/(2951.9KJ/kg-162.1KJ/Kg)
η=0.3=30%
