Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
k = 1.4
Work done is given as;
inlet velocity is negligible;
Therefore, the exit velocity is 629.41 m/s
Answer:
1/f = 1/i + 1/o thin lens equation
1/i = 1/f - 1/o
i = o * f / ( o - f) = 50 * (-20) / (50 - (-20)) = -14.3 cm
The final image is erect and 14.3 cm behind the curved surface
M = -o / i = 14.3 / 50 = .29 magnificaton of object
S = .29 * 25 cm = 7.1 cm appearance of bird in mirror (height)
The two forces should be equal therefore:
2.10 * Fa = Fa + 2 * F * cos 18
simplifying the right side:
2.10 * Fa = Fa + 1.902 * F
1.10 Fa = 1.902 F
<span>F / Fa = 0.578</span>
The only thing we know of so far that can shift light to longer wavelengths is the "Doppler" effect. If the source and the observer are moving apart, then the observer sees wavelengths that are longer than they should be. If the source and the observer are moving toward each other, then the observer sees wavelengths that are shorter than they should be. It works for ANY wave ... sound, light, water etc. The trick is to know what the wavelength SHOULD be. If you know that, then you can tell whether you and the source are moving together or apart, and you can even tell how fast. If the lines in a star"s spectrum are at wavelengths that are too long, then from everything we know right now, the star and Earth are moving apart.
Answer:
V = 6.53 × 10^14
h = 6.626 × 10^-34
E = hV
E = (6.626 × 10^-34) (6.53 × 10^14)
E = 4.33 × 10^-19
