The refractive index of water is
![n=1.33](https://tex.z-dn.net/?f=n%3D1.33)
. This means that the speed of the light in the water is:
![v= \frac{c}{n}= \frac{3 \cdot 10^8 m/s}{1.33 }=2.26 \cdot 10^8 m/s](https://tex.z-dn.net/?f=v%3D%20%5Cfrac%7Bc%7D%7Bn%7D%3D%20%5Cfrac%7B3%20%5Ccdot%2010%5E8%20m%2Fs%7D%7B1.33%20%7D%3D2.26%20%5Ccdot%2010%5E8%20m%2Fs%20%20)
The relationship between frequency f and wavelength
![\lambda](https://tex.z-dn.net/?f=%5Clambda)
of a wave is given by:
![\lambda= \frac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%3D%20%5Cfrac%7Bv%7D%7Bf%7D%20)
where v is the speed of the wave in the medium. The frequency of the light does not change when it moves from one medium to the other one, so we can compute the ratio between the wavelength of the light in water
![\lambda_w](https://tex.z-dn.net/?f=%5Clambda_w)
to that in air
![\lambda](https://tex.z-dn.net/?f=%5Clambda)
as
![\frac{\lambda_w}{\lambda}= \frac{ \frac{v}{f} }{ \frac{c}{f} } = \frac{v}{c}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Clambda_w%7D%7B%5Clambda%7D%3D%20%5Cfrac%7B%20%5Cfrac%7Bv%7D%7Bf%7D%20%7D%7B%20%5Cfrac%7Bc%7D%7Bf%7D%20%7D%20%3D%20%5Cfrac%7Bv%7D%7Bc%7D%20)
where v is the speed of light in water and c is the speed of light in air. Re-arranging this formula and by using
![\lambda=400 nm](https://tex.z-dn.net/?f=%5Clambda%3D400%20nm)
, we find
![\lambda_w = \lambda \frac{v}{c}=(400 nm) \frac{2.26 \cdot 10^8 m/s}{3 \cdot 10^8 m/s}=301 nm](https://tex.z-dn.net/?f=%5Clambda_w%20%3D%20%5Clambda%20%5Cfrac%7Bv%7D%7Bc%7D%3D%28400%20nm%29%20%5Cfrac%7B2.26%20%5Ccdot%2010%5E8%20m%2Fs%7D%7B3%20%5Ccdot%2010%5E8%20m%2Fs%7D%3D301%20nm%20)
which is the wavelength of light in water.
Answer:
The net Electric field at the mid point is 289.19 N/C
Given:
Q = + 71 nC = ![71\times 10^{- 9} C](https://tex.z-dn.net/?f=71%5Ctimes%2010%5E%7B-%209%7D%20C)
Q' = + 42 nC = ![42\times 10^{- 9} C](https://tex.z-dn.net/?f=42%5Ctimes%2010%5E%7B-%209%7D%20C)
Separation distance, d = 1.9 m
Solution:
To find the magnitude of electric field at the mid point,
Electric field at the mid-point due to charge Q is given by:
![\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cfrac%7BQ%7D%7B4%5Cpi%5Cepsilon_%7Bo%7D%28%5Cfrac%7Bd%7D%7B2%7D%29%5E%7B2%7D%7D)
![\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B71%5Ctimes%2010%5E%7B-%209%7D%7D%7B4%5Cpi%5C8.85%5Ctimes%2010%5E%7B-%2012%7D%28%5Cfrac%7B1.9%7D%7B2%7D%29%5E%7B2%7D%7D)
![\vec{E} = 708.03 N/C](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20708.03%20N%2FC)
Now,
Electric field at the mid-point due to charge Q' is given by:
![\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BE%27%7D%20%3D%20%5Cfrac%7BQ%27%7D%7B4%5Cpi%5Cepsilon_%7Bo%7D%28%5Cfrac%7Bd%7D%7B2%7D%29%5E%7B2%7D%7D)
![\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BE%27%7D%20%3D%20%5Cfrac%7B42%5Ctimes%2010%5E%7B-%209%7D%7D%7B4%5Cpi%5C8.85%5Ctimes%2010%5E%7B-%2012%7D%28%5Cfrac%7B1.9%7D%7B2%7D%29%5E%7B2%7D%7D)
![\vec{E'} = 418.84 N/C](https://tex.z-dn.net/?f=%5Cvec%7BE%27%7D%20%3D%20418.84%20N%2FC)
Now,
The net Electric field is given by:
![\vec{E_{net}} = \vec{E} - \vec{E'}](https://tex.z-dn.net/?f=%5Cvec%7BE_%7Bnet%7D%7D%20%3D%20%5Cvec%7BE%7D%20-%20%5Cvec%7BE%27%7D)
![\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C](https://tex.z-dn.net/?f=%5Cvec%7BE_%7Bnet%7D%7D%20%3D%20708.03%20-%20418.84%20%3D%20289.19%20N%2FC)
Answer:
Work done,W= 250J
Displacement , s = 60
We know that, Work done = Force x displacement
i.e , W = Fxs
250 J = F x 60m
F = 250/60
=4.16 N
Hence , 4.16 N of Force is applied on the body.
If one of the answers are the coin will float that one would be correct.
Answer:
A: Soil
Explanation:
Protists need a moist environment to survive, and shallow ponds, oceans, and blood is all moist. So, the answer would be the soil, because that is the least moist environment out of these options.