Answer:
V2 = 15.53 [m/s]]
Explanation:
In order to solve this problem we must use the principle of energy conservation, where potential energy is transformed into kinetic energy. At the bottom is taken as a reference level of potential energy, where the value of this energy is equal to zero.
Above the inclined plane we have two energies, kinetics and potential. While when the sled is at the reference level all this energy will have been transformed into kinetic energy.
![E_{1}=E_{2}\\ m*g*h+(\frac{1}{2} )*m*v_{1} ^{2}=\frac{1}{2}*m*v_{2} ^{2} \\(9.81*8.21)+(0.5*8.96^{2} )=(0.5*v_{2}^{2} )\\(0.5*v_{2}^{2} )=120.68\\v_{2} ^{2}=241.36\\v_{2} =\sqrt{241.36}\\ v_{2} =15.53[m/s]](https://tex.z-dn.net/?f=E_%7B1%7D%3DE_%7B2%7D%5C%5C%20%20m%2Ag%2Ah%2B%28%5Cfrac%7B1%7D%7B2%7D%20%29%2Am%2Av_%7B1%7D%20%5E%7B2%7D%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2Av_%7B2%7D%20%5E%7B2%7D%20%5C%5C%289.81%2A8.21%29%2B%280.5%2A8.96%5E%7B2%7D%20%29%3D%280.5%2Av_%7B2%7D%5E%7B2%7D%20%29%5C%5C%280.5%2Av_%7B2%7D%5E%7B2%7D%20%29%3D120.68%5C%5Cv_%7B2%7D%20%5E%7B2%7D%3D241.36%5C%5Cv_%7B2%7D%20%3D%5Csqrt%7B241.36%7D%5C%5C%20%20v_%7B2%7D%20%3D15.53%5Bm%2Fs%5D)
There is no displacement. The frog is back where it began.
Answer:
the pressure fluctuation is LONGITUDINAL
Explanation:
Sound waves are an oscillating movement of air particles, this can be analyzed in two different, but equivalent ways, as an air oscillation and with a pressure wave due to these oscillations.
The expression for the wave is
ΔP = Δo sin (kx - wt)
Therefore, the pressure variation is in the same direction as the displacement variation, consequently the pressure fluctuation is LONGITUDINAL
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