Answer:4.93 m/s
Explanation:
Given
height to reach is (h )1.24 m
here Let initial velocity is u
using equation of motion
![v^2-u^2=2ah](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2ah)
here Final Velocity v=0
a=acceleration due to gravity
![0-u^2=2\left ( -g\right )h](https://tex.z-dn.net/?f=0-u%5E2%3D2%5Cleft%20%28%20-g%5Cright%20%29h)
![u=\sqrt{2gh}](https://tex.z-dn.net/?f=u%3D%5Csqrt%7B2gh%7D)
![u=\sqrt{2\times 9.81\times 1.24}](https://tex.z-dn.net/?f=u%3D%5Csqrt%7B2%5Ctimes%209.81%5Ctimes%201.24%7D)
![u=\sqrt{24.328}](https://tex.z-dn.net/?f=u%3D%5Csqrt%7B24.328%7D)
u=4.93 m/s
Answer:
So logically speaking the question you have asked makes no sense
Explanation:
Zero what? After landing where? How what? before asking a question think about how other people will read it.
Answer:
![E = 6.77 *10^{3} N/C](https://tex.z-dn.net/?f=E%20%3D%206.77%20%2A10%5E%7B3%7D%20N%2FC)
Explanation:
THE GIVEN sheet can be taken as two horizontal force with surface charge density is
at one surface is ∈_1 = ![95*10^ {-9} C/mm2](https://tex.z-dn.net/?f=95%2A10%5E%20%7B-9%7D%20C%2Fmm2)
at oher surface is ∈_2= ![-25*10^{-9} C/mm2](https://tex.z-dn.net/?f=-25%2A10%5E%7B-9%7D%20C%2Fmm2)
the magnitude of electric field due to surface charge is given as![\frac{∈}{2∈_O}](https://tex.z-dn.net/?f=%5Cfrac%7B%E2%88%88%7D%7B2%E2%88%88_O%7D)
So, electric field at P (2 CM below from surface is) = E_1 +E_2
![E = \frac{∈_1}{2∈_O} +\frac{∈_2}{2∈_O}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B%E2%88%88_1%7D%7B2%E2%88%88_O%7D%20%2B%5Cfrac%7B%E2%88%88_2%7D%7B2%E2%88%88_O%7D)
![E = \frac{95*10^{-9} +25*10^{-9}}{2*8.85*10^{-12}}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B95%2A10%5E%7B-9%7D%20%2B25%2A10%5E%7B-9%7D%7D%7B2%2A8.85%2A10%5E%7B-12%7D%7D)
![E = 6.77 *10^{3} N/C](https://tex.z-dn.net/?f=E%20%3D%206.77%20%2A10%5E%7B3%7D%20N%2FC)
It doubles
Momentum= mass * velocity
when velocity doubles
momentum= 2*(velocity* mass)
Answer:
15.106 N
Explanation:
From the given information,
The weight of the bucket can be calculated as:
![W_b = m_bg = \\ \\ W_b = (0.730 \ kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N](https://tex.z-dn.net/?f=W_b%20%3D%20m_bg%20%3D%20%20%5C%5C%20%5C%5C%20%20W_b%20%3D%20%280.730%20%5C%20%20kg%29%20%28%209.80%20%5C%20m%2Fs%5E2%29%20%5C%5C%20%5C%5C%20W_b%20%3D%207.154%20%5C%20N)
The mass of the water accumulated in the bucket after 3.20s is:
![m_w= (0.20 \ L/s) ( 3.20)s](https://tex.z-dn.net/?f=m_w%3D%20%280.20%20%5C%20L%2Fs%29%20%28%203.20%29s)
![m _w=0.64 \ kg](https://tex.z-dn.net/?f=m%20_w%3D0.64%20%5C%20kg)
To determine the weight of the water accumulated in the bucket, we have:
![W_w = m_w g](https://tex.z-dn.net/?f=W_w%20%3D%20m_w%20g)
![W_w = ( 0.64 \ kg )(9.80\ m \ /s^2)](https://tex.z-dn.net/?f=W_w%20%3D%20%28%200.64%20%20%5C%20kg%20%29%289.80%5C%20%20m%20%20%5C%20%20%2Fs%5E2%29)
![W_w = 6.272 \ N](https://tex.z-dn.net/?f=W_w%20%3D%206.272%20%5C%20N)
For the speed of the water before hitting the bucket; we have:
![v = \sqrt{2gh}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2gh%7D)
![v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2%2A9.80%20%5C%20m%2Fs%5E2%20%2A%203.60%20%5C%20m%7D)
v = 8.4 m/s
Now, the force required to stop the water later when it already hit the bucket is:
![F = v ( \dfrac {dm}{dt} )](https://tex.z-dn.net/?f=F%20%3D%20v%20%28%20%5Cdfrac%20%7Bdm%7D%7Bdt%7D%20%29)
![F = (8.4 \ m/s)( 0.200 \ L/s)](https://tex.z-dn.net/?f=F%20%3D%20%288.4%20%5C%20m%2Fs%29%28%200.200%20%5C%20L%2Fs%29)
F = 1.68 N
Finally, the reading scale is:
= 7.154 N + 6.272 N + 1.68 N
= 15.106 N