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dedylja [7]
3 years ago
13

Orbit is to _____ as altitude is to _____.

Engineering
2 answers:
Brrunno [24]3 years ago
7 0

Answer:

Path/Distance

Explanation:

on edg

lubasha [3.4K]3 years ago
5 0

Answer:

What is the minimum altitude for orbit?

Technically, objects in low-Earth orbit are at an altitude of between 160 to 2,000 km (99 to 1200 mi) above the Earth's surface. Any object below this altitude will being to suffer from orbital decay and will rapidly descend into the atmosphere, either burning up or crashing on the surface.

Explanation:

orbital-mechanics orbit low-earth-orbit altitude. The altitude of a satellite is the distance between the Earth's surface and the satellite, but the Earth itself is not spherical. At the equator the Earth's radius is 21 km more than at the poles, and in fact the shape of the Earth is not even a perfect oblate spheroid.A Japan Aerospace Exploration Agency (JAXA) satellite has been certified by Guiness World Records as setting a new altitude orbit record. However, while most satellites orbit high above the Earth, the JAXA satellite Tsubame was recognized for achieving the lowest altitude, of only 167.4 kilometers.

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Air enters an adiabatic compressor through 0:5m2 opening and exhausts through a 0:2m2 opening. The inlet conditions of air are 2
gladu [14]

Answer:

i) 43.55 kg/s

ii) 40 m/s

iii)  -199.32 KW

Explanation:

To resolve the above question we have to make some assumptions :

  • mass flow  through the system is constant
  • The only interactions that are between the system and the surrounding are work and heat
  • The fluid is uniform

i) first we have to determine the mass flow rate of the air

M = (\frac{P1}{RT1} )A1v1

    = ( \frac{200}{0.287*325} ) 0.5 * V1 ---------- (1) hence  M = 43.55 kg/s

ii) using this relationship : A1V1 = A2V2 hence V1 = (0.2/0.5) * 100 = 40m/s (    inlet velocity )

input this value into equation 1

iii) Next we will determine the power required to run compressor

attached below

power required = -199.32 KW ( this value indicates that there is power supplied )

8 0
3 years ago
1. A flywheel is suspended by resting the inside of the rim on a horizontal knife edge so that the wheel can swing in a vertical
sammy [17]

Answer: A fly wheel having a mass of 30kg was allowed to swing as pendulum about a knife edge at inner side of the rim as shown in figure.

Explanation:

8 0
3 years ago
Using the formula XC=1/(2πfC) in your answer, how would a capacitor influence a simple DC series circuit?
Kisachek [45]

Answer:

jdjdbddifnifndjrnirnrinr

5 0
3 years ago
FAULT LOCATION METHODS(input-output)<br>​
ValentinkaMS [17]

Fault location techniques are used in power systems for accurate pinpointing of the fault position.

This paper presents a comparative study between two fault location methods in distribution network with Distributed Generation (DG). Both methods are based on computing the impedance using fundamental voltage and current signals. The first method uses one-end information and the second uses both ends

6 0
3 years ago
What is the thermal efficiency of this reheat cycle in terms of enthalpies?
schepotkina [342]

Answer:

   \eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}

Explanation:

For close gas turbine:

       Gas turbine works on Brayton cycle.Gas turbine have lots of applications like ,it is use in aircraft,in land applications etc.

Reheating is the method to improve the efficiency of the gas turbine.In reheating gas is expanding in two turbine instead of one turbine alone.Two turbine like high pressure turbine and low pressure turbine are used for expansion.

In the above diagram 1-2 is a compressor,2-3 heat addition,3-4 high pressure turbine,4-5 reheating of cycle 5-6 low pressure turbine,6-1 heat rejection,

We know that    \eta =\frac{W_{net}}{Q_{s}}

Now take h_{1},h_{2},,h_{3},h_{4},h_{5},h_{6} represent the enthalpy of point 1,2,3,4,5,6 in the cycle respectively.

So total heat supplied Q_S=

\left (h_3-h_2\right )+\left (h_5-h_4\right )

Net work out put

W_{net}=\left (h_5-h_6\right )-\left (h_2-h_1\right )

So efficiency   \eta =\frac{W_{net}}{Q_{s}}

      \eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}

5 0
4 years ago
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