Answer:
![1.8\times 105 N/C](https://tex.z-dn.net/?f=1.8%5Ctimes%20105%20N%2FC)
Explanation:
We are given that
![u=2\times 10^7 m/s](https://tex.z-dn.net/?f=u%3D2%5Ctimes%2010%5E7%20m%2Fs)
![v=4\times 10^7 m/s](https://tex.z-dn.net/?f=v%3D4%5Ctimes%2010%5E7%20m%2Fs)
d=1.9 cm=![\frac{1.9}{100}=0.019 m](https://tex.z-dn.net/?f=%5Cfrac%7B1.9%7D%7B100%7D%3D0.019%20m)
Using 1m=100 cm
We have to find the electric field strength.
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
Using the formula
![(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)](https://tex.z-dn.net/?f=%284%5Ctimes%2010%5E7%29%5E2-%282%5Ctimes%2010%5E7%29%5E2%3D2a%280.019%29)
![16\times 10^{14}-4\times 10^{14}=0.038a](https://tex.z-dn.net/?f=16%5Ctimes%2010%5E%7B14%7D-4%5Ctimes%2010%5E%7B14%7D%3D0.038a)
![0.038a=12\times 10^{14}](https://tex.z-dn.net/?f=0.038a%3D12%5Ctimes%2010%5E%7B14%7D)
![a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B12%7D%7B0.038%7D%5Ctimes%2010%5E%7B14%7D%3D3.16%5Ctimes%2010%5E%7B16%7Dm%2Fs%5E2)
![q=1.6\times 10^{-19} C](https://tex.z-dn.net/?f=q%3D1.6%5Ctimes%2010%5E%7B-19%7D%20C)
Mass of electron,m![=9.1\times 10^{-31} kg](https://tex.z-dn.net/?f=%3D9.1%5Ctimes%2010%5E%7B-31%7D%20kg)
![E=\frac{ma}{q}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bma%7D%7Bq%7D)
Substitute the values
![E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B9.1%5Ctimes%2010%5E%7B-31%7D%5Ctimes%203.16%5Ctimes%2010%5E%7B16%7D%7D%7B1.6%5Ctimes%2010%5E%7B-19%7D%7D)
![E=1.8\times 105 N/C](https://tex.z-dn.net/?f=E%3D1.8%5Ctimes%20105%20N%2FC)
S=Vt
110=V(72)
110/72=V
V=1.527m/s
Answer:
Time period of the motion will remain the same while the amplitude of the motion will change
Explanation:
As we know that time period of oscillation of spring block system is given as
![T= 2\pi\sqrt{\frac{M}{k}}](https://tex.z-dn.net/?f=T%3D%202%5Cpi%5Csqrt%7B%5Cfrac%7BM%7D%7Bk%7D%7D)
now we know that
M = mass of the object
k = spring constant
So here we know that the time period is independent of the gravity
while the maximum displacement of the spring from its mean position will depends on the gravity as
![mg = kx](https://tex.z-dn.net/?f=mg%20%3D%20kx)
![x = \frac{mg}{k}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7Bmg%7D%7Bk%7D)
so we can say that
Time period of the motion will remain the same while the amplitude of the motion will change
Answer: ![0.75\ cm](https://tex.z-dn.net/?f=0.75%5C%20cm)
Explanation:
Given
Wavelength of light ![\lambda=500\ nm](https://tex.z-dn.net/?f=%5Clambda%3D500%5C%20nm)
Screen is
away
Distance between two adjacent bright fringe is ![\Delta y=\dfrac{\lambda D}{d}](https://tex.z-dn.net/?f=%5CDelta%20y%3D%5Cdfrac%7B%5Clambda%20D%7D%7Bd%7D)
When same experiment done in water, wavelength reduce to ![\dfrac{\lambda }{\mu}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Clambda%20%7D%7B%5Cmu%7D)
So, the distance between the two adjacent bright fringe is ![\Delta y'=\dfrac{\lambda D}{\mu d}](https://tex.z-dn.net/?f=%5CDelta%20y%27%3D%5Cdfrac%7B%5Clambda%20D%7D%7B%5Cmu%20d%7D)
Keeping other factor same, distance becomes
![\Rightarrow \dfrac{1}{\frac{4}{3}}=\dfrac{3}{4}\quad \text{Refractive index of water is }\dfrac{4}{3}\\\\\Rightarrow 0.75\ cm](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cdfrac%7B1%7D%7B%5Cfrac%7B4%7D%7B3%7D%7D%3D%5Cdfrac%7B3%7D%7B4%7D%5Cquad%20%5Ctext%7BRefractive%20index%20of%20water%20is%20%7D%5Cdfrac%7B4%7D%7B3%7D%5C%5C%5C%5C%5CRightarrow%200.75%5C%20cm)
No
It means the resultant electrical charges had canceled each other out so there is no field to be sensed .