Answer:
before wheel is put on it should be looked at for damage and a sound or ring test should be done to check for cracks, to test the wheel it should be tapped with a non metallic instrument (I looked it up)
Answer:
The force induced on the aircraft is 2.60 N
Solution:
As per the question:
Power transmitted, ![P_{t} = 8 kW = 8000 W](https://tex.z-dn.net/?f=P_%7Bt%7D%20%3D%208%20kW%20%3D%208000%20W)
Now, the force, F is given by:
(1)
where
v = velocity
Now,
For a geo-stationary satellite, the centripetal force,
is provided by the gravitational force,
:
![F_{c} = F_{G}](https://tex.z-dn.net/?f=F_%7Bc%7D%20%3D%20F_%7BG%7D)
![\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E%7B2%7D%7D%7BR%7D%20%3D%20%5Cfrac%7BGM_%7Be%7Dm%7BR%5E%7B2%7D%7D)
Thus from the above, velocity comes out to be:
![v = \sqrt{\frac{GM_{e}}{R}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7BGM_%7Be%7D%7D%7BR%7D%7D)
![v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.979\times 10^{24}}{42166\times 10^{3}}} = 3075.36 m/ s](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B6.67%5Ctimes%2010%5E%7B-%2011%7D%5Ctimes%205.979%5Ctimes%2010%5E%7B24%7D%7D%7B42166%5Ctimes%2010%5E%7B3%7D%7D%7D%20%3D%203075.36%20m%2F%20s)
where
R = ![R_{e} + H](https://tex.z-dn.net/?f=R_%7Be%7D%20%2B%20H)
R = ![\sqrt{GM_{e}(\frac{T}{2\pi})^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7BGM_%7Be%7D%28%5Cfrac%7BT%7D%7B2%5Cpi%7D%29%5E%7B2%7D%7D)
where
G = Gravitational constant
T = Time period of rotation of Earth
R is calculated as 42166 km
Now, from eqn (1):
![8000 = F\times 3075.36](https://tex.z-dn.net/?f=8000%20%3D%20F%5Ctimes%203075.36)
F = 2.60 N
Answer:
a) 0.489
b) 54.42 kg/s
c) 247.36 kW/s
Explanation:
Note that all the initial enthalpy and entropy values were gotten from the tables.
See the attachment for calculations
Answer:
There is 0.466 KW required to operate this air-conditioning system
Explanation:
<u>Step 1:</u> Data given
Heat transfer rate of the house = Ql = 755 kJ/min
House temperature = Th = 24°C = 24 +273 = 297 Kelvin
Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin
<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.
COPr,c = 1 / ((To/Th) - 1)
COPr,c = 1 /(( 308/297) - 1)
COPr,c = 1/ 0.037
COPr,c = 27
<u>Step 3:</u> The power input cna be given as followed:
Wnet,in = Ql / COPr,max
Wnet, in = 755 / 27
Wnet,in = 27.963 kJ/min
Win = 27.963 * 1 KW/60kJ/min = 0.466 KW
There is 0.466 KW required to operate this air-conditioning system