Answer:
a
The number of fringe is z = 3 fringes
b
The ratio is 
Explanation:
a
From the question we are told that
The wavelength is 
The distance between the slit is 
The width of the slit is 
let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

Substituting values
z = 3 fringes
b
From the question we are told that the order of the bright fringe is n = 3
Generally the intensity of a pattern is mathematically represented as
![I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%5B%5Cfrac%7B%5Cpi%20d%20sin%20%5Ctheta%7D%7B%5Clambda%7D%20%5D%5B%5Cfrac%7Bsin%20%28%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%20%7D%20%29%7D%7B%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%7D%20%7D%20%5D)
Where
is the intensity of the central fringe
And Generally 
![I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20co%5E2%20%5B%20%5Cfrac%7B%5Cpi%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%5D%20%5B%5Cfrac%7B%5Cfrac%7Bsin%20%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%7D%7B%5Cfrac%7B%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%28n%20%5Cpi%29%5B%5Cfrac%7B%5Cfrac%7Bsin%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%29%7D%7B%20%5Cfrac%7B%20%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%20%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%283%20%5Cpi%29%20%5B%5Cfrac%7Bsin%20%28%5Cfrac%7B3%20%5Cpi%20%7D%7B6%7D%20%29%7D%7B%5Cfrac%7B3%20%5Cpi%7D%7B6%7D%20%7D%20%5D)


Answer:
1m/s is the acceleration used. C
Explanation:
please mark brainliest
A) gases particles are far apart
Answer:
The time taken is 
Explanation:
From the question we are told that
The length of steel the wire is 
The length of the copper wire is 
The diameter of the wire is 
The tension is 
The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

Where
is the time taken to transverse the steel wire which is mathematically represented as
![t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]](https://tex.z-dn.net/?f=t_s%20%20%3D%20l_1%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B%5Crho%20%2A%20%5Cpi%20%2A%20%20d%5E2%20%7D%7B%204%20%2A%20%20T%7D%20%7D%20%5D)
here
is the density of steel with a value 
So
![t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]](https://tex.z-dn.net/?f=t_s%20%20%3D%2031%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B8920%20%2A%203.142%2A%20%20%281%2A10%5E%7B-3%7D%29%5E2%20%7D%7B%204%20%2A%20%20122%7D%20%7D%20%5D)

And
is the time taken to transverse the copper wire which is mathematically represented as
![t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]](https://tex.z-dn.net/?f=t_c%20%20%3D%20l_2%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B%5Crho_c%20%2A%20%5Cpi%20%2A%20%20d%5E2%20%7D%7B%204%20%2A%20%20T%7D%20%7D%20%5D)
here
is the density of steel with a value 
So
![t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]](https://tex.z-dn.net/?f=t_c%20%20%3D%2017%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B7860%20%2A%203.142%2A%20%20%281%2A10%5E%7B-3%7D%29%5E2%20%7D%7B%204%20%2A%20%20122%7D%20%7D%20%5D)

So


