1250 J in 5 sec= 250 Joule(s) per second (1250/5 0
250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)
250 Watts output = 250/0.65 efficiency = 384 Watts input
1 Horsepower = 732 Watts
Motors 1 Horsepower and under are made in certain step sizes like
3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.
3/4 Horsepower is 549 Watts
1/2 Horsepower is 366 Watts
so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.
ANSWER: Due to the propulsion from the inlet diameter of this engine bring 1.6 m allows the compressor rations to radiate allowing thrust propultion above all velocitic rebisomes.
We can calculate this with the law of conservation of energy. Here we have a food package with a mass m=40 kg, that is in the height h=500 m and all of it's energy is potential. When it is dropped, it's potential energy gets converted into kinetic energy. So we can say that its kinetic and potential energy are equal, because we are neglecting air resistance:
Ek=Ep, where Ek=(1/2)*m*v² and Ep=m*g*h, where m is the mass of the body, g=9.81 m/s² and h is the height of the body.
(1/2)*m*v²=m*g*h, masses cancel out and we get:
(1/2)*v²=g*h, and we multiply by 2 both sides of the equation
v²=2*g*h, and we take the square root to get v:
v=√(2*g*h)
v=99.04 m/s
So the package is moving with the speed of v= 99.04 m/s when it hits the ground.
Answer:
The kinetic energy is: 50[J]
Explanation:
The ball is having a potential energy of 100 [J], therefore
PE = [J]
The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.
![E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%3D%20gravity%5Bm%2Fs%5E%7B2%7D%20%5D%5C%5Cm%20%3D%20mass%20%5Bkg%5D%5C%5Cm%3D%20%5Cfrac%7BE_%7Bp%7D%20%7D%7Bg%2Ah%7D%5C%5C%20m%3D%20%5Cfrac%7B100%7D%7B9.81%2A10%7D%5C%5C%5C%5Cm%3D%201.01%5Bkg%5D%5C%5C%5C%5C)
In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.
When the elevation is 5 [m], we have a potential energy of
![P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\](https://tex.z-dn.net/?f=P_%7Be%7D%20%3Dm%2Ag%2Ah%5C%5CP_%7Be%7D%20%3D1.01%2A9.81%2A5%5C%5C%5C%5CP_%7Be%7D%20%3D%2050%20%5BJ%5D%5C%5C)
This energy is equal to the kinetic energy, therefore
Ke= 50 [J]
