Answer:
v = 12.4 [m/s]
Explanation:
With the speed and Area information, we can determine the volumetric flow.
where:
r = radius = 0.0120 [m]
v = 2.88 [m/s]
![A=\pi *(0.0120)^{2} \\A=4.523*10^{-4} [m]\\](https://tex.z-dn.net/?f=A%3D%5Cpi%20%2A%280.0120%29%5E%7B2%7D%20%5C%5CA%3D4.523%2A10%5E%7B-4%7D%20%5Bm%5D%5C%5C)
Therefore the flow is:
![V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]](https://tex.z-dn.net/?f=V%3D2.88%2A4.523%2A10%5E%7B-4%7D%20%5C%5CV%3D1.302%2A10%5E%7B-3%7D%20%5Bm%5E%7B3%7D%2Fs%20%5D)
Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.
![v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]](https://tex.z-dn.net/?f=v%3DV%2FA%5C%5Cv%3D1.302%2A10%5E%7B-3%7D%20%2F1.05%2A10%5E%7B-4%7D%20%5C%5Cv%3D12.4%5Bm%2Fs%5D)
Answer:
B
Explanation:
Because it has to increase
False............ It is in both directions in the form of centri-petal & fugal towards inwards and outwards respectively....
<h2>
The impact is occurred at a distance 348.55 m far from person.</h2>
Explanation:
Let s be the distance to the impact position and t be the time when he hears the sound though concrete after impact.
Time when he hears impact through air = t + 0.90
The distance traveled by sound wave in both concrete and air is s.
Speed of sound in air = 343 m/s
Speed of sound in concrete = 3000 m/s
Distance traveled in air = Distance traveled in concrete
343 x Time when he hears impact through air = 3000 x Time when he hears impact through concrete
343 x (t + 0.90) = 3000 x t
3000t - 343 t = 343 x 0.90
2657t = 308.7
t = 0.116 s
Distance traveled in concrete = 3000 x t = 3000 x 0.116 = 348.55m
So the impact is occurred at a distance 348.55 m far from person.
