Answer:
N = 38546.82 rpm
Explanation:
= 150 mm
![A_{1}= \frac{\pi }{4}\times 150^{2}](https://tex.z-dn.net/?f=A_%7B1%7D%3D%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Ctimes%20150%5E%7B2%7D)
= 17671.45 ![mm^{2}](https://tex.z-dn.net/?f=mm%5E%7B2%7D)
= 250 mm
![A_{2}= \frac{\pi }{4}\times 250^{2}](https://tex.z-dn.net/?f=A_%7B2%7D%3D%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Ctimes%20250%5E%7B2%7D)
= 49087.78 ![mm^{2}](https://tex.z-dn.net/?f=mm%5E%7B2%7D)
The centrifugal force acting on the flywheel is fiven by
F = M (
-
) x
------------(1)
Here F = ( -UTS x
+ UCS x
)
Since density, ![\rho = \frac{M}{V}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7BM%7D%7BV%7D)
![\rho = \frac{M}{A\times t}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7BM%7D%7BA%5Ctimes%20t%7D)
![M = \rho \times A\times t](https://tex.z-dn.net/?f=M%20%3D%20%5Crho%20%5Ctimes%20A%5Ctimes%20t)
![M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t](https://tex.z-dn.net/?f=M%20%3D%207100%20%5Ctimes%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Cleft%20%28%20D_%7B2%7D%5E%7B2%7D-D_%7B1%7D%5E%7B2%7D%20%5Cright%20%29%5Ctimes%20t)
![M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37](https://tex.z-dn.net/?f=M%20%3D%207100%20%5Ctimes%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Cleft%20%28%20250%5E%7B2%7D-150%5E%7B2%7D%20%5Cright%20%29%5Ctimes%2037)
![M = 8252963901](https://tex.z-dn.net/?f=M%20%3D%208252963901)
∴
-
= 50 mm
∴ F = ![763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}](https://tex.z-dn.net/?f=763%5Ctimes%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Ctimes%20250%5E%7B2%7D-217%5Ctimes%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Ctimes%20150%5E%7B2%7D)
F = 33618968.38 N --------(2)
Now comparing (1) and (2)
![33618968.38 = 8252963901\times 50\times \omega ^{2}](https://tex.z-dn.net/?f=33618968.38%20%3D%208252963901%5Ctimes%2050%5Ctimes%20%5Comega%20%5E%7B2%7D)
∴ ω = 4036.61
We know
![\omega = \frac{2\pi N}{60}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B2%5Cpi%20N%7D%7B60%7D)
![4036.61 = \frac{2\pi N}{60}](https://tex.z-dn.net/?f=4036.61%20%3D%20%5Cfrac%7B2%5Cpi%20N%7D%7B60%7D)
∴ N = 38546.82 rpm
Okay sure.
I’ll 1)chords
2)pulse
3)aerophone
4) the answer is C
5)rhythm
Pretty sure those are the answers
Answer:
![P_2-P_1=27209h](https://tex.z-dn.net/?f=P_2-P_1%3D27209h)
Explanation:
For pressure gage we can determine this by saying:
The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.
![P_1+p_1gh_1=p_2_gh_2+P_2](https://tex.z-dn.net/?f=P_1%2Bp_1gh_1%3Dp_2_gh_2%2BP_2)
We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h
![P_2-P_1=27209h](https://tex.z-dn.net/?f=P_2-P_1%3D27209h)
If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will: a) increase during a climb.
<h3>What is a
ram air input?</h3>
A ram air input can be defined as an air intake system which is designed and developed to use the dynamic air pressure that is created due to vehicular motion, or ram pressure, in order to increase the static air pressure within the intake manifold of an internal combustion engine of an automobile.
This ultimately implies that, a ram air input allows a greater mass-flow of air through the engine of an automobile, thereby, increasing the engine's power.
In conclusion, indicated airspeed will increase during a climb when both the ram air input and drain hole of the pitot system become blocked.
Read more on pilots here: brainly.com/question/10381526
#SPJ1
Complete Question:
If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will
a) increase during a climb
b) decrease during a climb
c) remain constant regardless of altitude change
The ratio between a and b is 1/3