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
I think that it is all of the above
Answer:
please mark me as a brainleast
Explanation:
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In Engineering, the thrust angle is checked by referencing: C. vehicle centerline.
<h3>What is a
thrust angle?</h3>
A thrust angle can be defined as an imaginary line which is drawn perpendicularly from the centerline of the rear axle of a vehicle, down the centerline.
This ultimately implies that, the thrust angle is a reference to the centerline (wheelbase) of a vehicle, and it confirms that the two wheels on both sides are properly angled within specification.
Read more on thrust angle here: brainly.com/question/13000914
#SPJ1
Answer:
s= 20.4 m
Explanation:
First lets write down equations for each ball:
s=so+vo*t+1/2a_c*t^2
for ball A:
s_a=30+5*t+1/2*9.81*t^2
for ball B:
s_b=20*t-1/2*9.81*t^2
to find time deeded to pass we just put that
s_a = s_b
30+5*t-4.91*t^2=20*t-4.9*t^2
t=2 s
now we just have to put that time in any of those equations an get distance from the ground:
s = 30 + 5*2 -1/2*9.81 *2^2
s= 20.4 m