Yes it will not hope this helps you can i get brainlist(-;
5). D
The bolt of lightning is a stream of electrons flowing from
a negative to a positive part of a cloud, or either up or down
between a charged cloud and the ground.
6). D
As the skateboard rolls down the ramp it loses potential energy and gains kinetic energy.
Find the answer in the attachment
Answer:
Velocity in the smaller pipe should not be included as an additional variable.
Explanation:
![$\Delta P = f(D_1, D_2, V, \rho, \mu)$](https://tex.z-dn.net/?f=%24%5CDelta%20P%20%3D%20f%28D_1%2C%20D_2%2C%20V%2C%20%5Crho%2C%20%5Cmu%29%24)
The dimensional formula of the variables are
![$\Delta P = FL^{-2} , D_1 = L, D_2=L, V=LT_{-1}, \rho =FL^{-4}T^2, \mu = FL^{-2}T$](https://tex.z-dn.net/?f=%24%5CDelta%20P%20%3D%20FL%5E%7B-2%7D%20%2C%20D_1%20%3D%20L%2C%20D_2%3DL%2C%20V%3DLT_%7B-1%7D%2C%20%5Crho%20%3DFL%5E%7B-4%7DT%5E2%2C%20%5Cmu%20%3D%20FL%5E%7B-2%7DT%24)
Now using Buchingham's Pi Theorem, 6 - 3 = 3 dimensional parameters are required.
Use, D_1, V, \mu as the repeating variables.
Therefore, ![$\pi = \Delta PD_1^aV^b \mu^c$](https://tex.z-dn.net/?f=%24%5Cpi%20%3D%20%5CDelta%20PD_1%5EaV%5Eb%20%5Cmu%5Ec%24)
![$(FL^{-2})(L)^a(LT^{-1})^b(FL^{-2}T)^c = F^0L^0T^0$](https://tex.z-dn.net/?f=%24%28FL%5E%7B-2%7D%29%28L%29%5Ea%28LT%5E%7B-1%7D%29%5Eb%28FL%5E%7B-2%7DT%29%5Ec%20%3D%20F%5E0L%5E0T%5E0%24)
From this
1+c=0
-2+a+b-2c=0
-b+c=0
c=-1, b = -1, a = 1
Now, ![$\pi_1=\frac{\Delta PD_1}{V\mu}=\frac{(ML^{-1}T^{-2})L}{(LT^{-1})(ML^{-1}T^{-1})}=M^0L^0T^0$](https://tex.z-dn.net/?f=%24%5Cpi_1%3D%5Cfrac%7B%5CDelta%20PD_1%7D%7BV%5Cmu%7D%3D%5Cfrac%7B%28ML%5E%7B-1%7DT%5E%7B-2%7D%29L%7D%7B%28LT%5E%7B-1%7D%29%28ML%5E%7B-1%7DT%5E%7B-1%7D%29%7D%3DM%5E0L%5E0T%5E0%24)
For ![$\pi_2 = D_2D_1^aV^b\mu^c$](https://tex.z-dn.net/?f=%24%5Cpi_2%20%3D%20D_2D_1%5EaV%5Eb%5Cmu%5Ec%24)
![$F^0L^0T^0=L(L)^a(LT^{-1})^b(FL^{-2}T)^c$](https://tex.z-dn.net/?f=%24F%5E0L%5E0T%5E0%3DL%28L%29%5Ea%28LT%5E%7B-1%7D%29%5Eb%28FL%5E%7B-2%7DT%29%5Ec%24)
c = 0
1 + a + b - 2c = 0
-b + c = 0
Therefore, a = -1, b = 0, c = 0
![$\pi_2 = \frac{D_2}{D_1}$](https://tex.z-dn.net/?f=%24%5Cpi_2%20%3D%20%5Cfrac%7BD_2%7D%7BD_1%7D%24)
For ![$\pi_3$](https://tex.z-dn.net/?f=%24%5Cpi_3%24)
![$\pi_3 = \rho, D_1^a V^b\mu^c$](https://tex.z-dn.net/?f=%24%5Cpi_3%20%3D%20%5Crho%2C%20D_1%5Ea%20V%5Eb%5Cmu%5Ec%24)
![$F^0L^0T^0 = (FL^{-4}T^2)(L)^a(LT^{-1})^b(FL^{-2}T)^c$](https://tex.z-dn.net/?f=%24F%5E0L%5E0T%5E0%20%3D%20%28FL%5E%7B-4%7DT%5E2%29%28L%29%5Ea%28LT%5E%7B-1%7D%29%5Eb%28FL%5E%7B-2%7DT%29%5Ec%24)
1 + c = 0
-4 + a + b - 2c = 0
2-b+c=0
c=-1, b=1, a = 1
Therefore, ![$\pi_3 = \frac{\rho D_1V}{\mu}$](https://tex.z-dn.net/?f=%24%5Cpi_3%20%20%3D%20%5Cfrac%7B%5Crho%20D_1V%7D%7B%5Cmu%7D%24)
Now checking,
![$\pi_3 = \frac{(ML^{-3})(L)(LT^{-1})}{ML^{-1}T^{-1}} = M^0L^0T^0$](https://tex.z-dn.net/?f=%24%5Cpi_3%20%3D%20%5Cfrac%7B%28ML%5E%7B-3%7D%29%28L%29%28LT%5E%7B-1%7D%29%7D%7BML%5E%7B-1%7DT%5E%7B-1%7D%7D%20%3D%20M%5E0L%5E0T%5E0%24)
Therefore, ![$\frac{\Delta P D_1}{V \mu} = \phi (\frac{D_2}{D_1}, \frac{\rho D_1 V}{\mu})$](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5CDelta%20P%20D_1%7D%7BV%20%5Cmu%7D%20%3D%20%5Cphi%20%28%5Cfrac%7BD_2%7D%7BD_1%7D%2C%20%5Cfrac%7B%5Crho%20D_1%20V%7D%7B%5Cmu%7D%29%24)
From continuity equation
![$V\frac{\pi}{4}D_1^2 = V_s \frac{\pi}{4}D_2^2$](https://tex.z-dn.net/?f=%24V%5Cfrac%7B%5Cpi%7D%7B4%7DD_1%5E2%20%3D%20V_s%20%5Cfrac%7B%5Cpi%7D%7B4%7DD_2%5E2%24)
![$V_s = V (\frac{D_1}{D_2})^2$](https://tex.z-dn.net/?f=%24V_s%20%3D%20V%20%28%5Cfrac%7BD_1%7D%7BD_2%7D%29%5E2%24)
is not independent of ![$D_1,D_2, V$](https://tex.z-dn.net/?f=%24D_1%2CD_2%2C%20V%24)
Therefore it should not be included as an additional variable.