Answer:
thats a lot, which one u want me to do?
Explanation:
To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.
The work done would be defined as
![\Delta W = \Delta PE + \Delta KE](https://tex.z-dn.net/?f=%5CDelta%20W%20%3D%20%5CDelta%20PE%20%2B%20%5CDelta%20KE)
Where,
PE = Potential Energy
KE = Kinetic Energy
![\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2](https://tex.z-dn.net/?f=%5CDelta%20W%20%3D%20%28%5CDelta%20m%29gh%2B%5Cfrac%7B1%7D%7B2%7D%28%5CDelta%20m%29v%5E2)
Where,
m = Mass
g = Gravitational energy
h = Height
v = Velocity
Considering power as the change of energy as a function of time we will then have to
![P = \frac{\Delta W}{\Delta t}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B%5CDelta%20W%7D%7B%5CDelta%20t%7D)
![P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B%5CDelta%20m%7D%7B%5CDelta%20t%7D%28gh%2B%5Cfrac%7B1%7D%7B2%7Dv%5E2%29)
The rate of mass flow is,
![\frac{\Delta m}{\Delta t} = \rho_w Av](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20m%7D%7B%5CDelta%20t%7D%20%3D%20%5Crho_w%20Av)
Where,
= Density of water
A = Area of the hose ![\rightarrow A=\pi r^2](https://tex.z-dn.net/?f=%5Crightarrow%20A%3D%5Cpi%20r%5E2)
The given radius is 0.83cm or
m, so the Area would be
![A = \pi (0.83*10^{-2})^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20%280.83%2A10%5E%7B-2%7D%29%5E2)
![A = 0.0002164m^2](https://tex.z-dn.net/?f=A%20%3D%200.0002164m%5E2)
We have then that,
![\frac{\Delta m}{\Delta t} = \rho_w Av](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20m%7D%7B%5CDelta%20t%7D%20%3D%20%5Crho_w%20Av)
![\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20m%7D%7B%5CDelta%20t%7D%20%3D%20%281000%29%280.0002164%29%285.4%29)
![\frac{\Delta m}{\Delta t} = 1.16856kg/s](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20m%7D%7B%5CDelta%20t%7D%20%3D%201.16856kg%2Fs)
Final the power of the pump would be,
![P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B%5CDelta%20m%7D%7B%5CDelta%20t%7D%28gh%2B%5Cfrac%7B1%7D%7B2%7Dv%5E2%29)
![P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)](https://tex.z-dn.net/?f=P%20%3D%20%281.16856%29%28%289.8%29%283.5%29%2B%5Cfrac%7B1%7D%7B2%7D5.4%5E2%29)
![P = 57.1192W](https://tex.z-dn.net/?f=P%20%3D%2057.1192W)
Therefore the power of the pump is 57.11W
Answer:longitudinal waves
Explanation:
They are longitudinal waves
Answer:
a bowling ball because it has the most mass.