Oh gosh oh I see it in my life face and
Answer:
The only work done is when the person lifts the sack over a distance, W = 78.48 [N]
Explanation:
We have to remember the definition of work, which tells us that work is the result of a force by a distance, we must apply this concept in each of the movements of the person in the problem described.
W = F * d
where:
F = force [N]
d = distance [m]
The force is given by the producto of the mass by the gravity.
F = 5 * 9.81 = 49.05 [N]
W = 49.05 * 1.6 = 78.48 [N]
Sound needs medium to travel and it can not travel without medium
so sound wave is a travelling wave
now we also know that sound wave propagate in form of rarefaction and compression.
So all medium particles travel in the direction of wave only
so it is a longitudinal wave also
so correct answer will be
<em>mechanical longitudinal </em>
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)
This equation is one of the most useful in classical physics. It is a concise statement of Isaac Newton's<span> Second Law of Motion, holding both the proportions and vectors of the Second Law. It translates as: The net force on an object is </span>equal<span> to the </span>mass<span>of the object multiplied by the </span>acceleration<span> of the object.</span>
