The Answer is Winds are caused by moving from air which is from high to low pressure
The answer is gravitational force. The gravitational force
between the earth and the moon is the similar as between any other two masses
in space.
Newton clarified that the force of attraction between two masses
is the outcome of the weight of object one multiplied by the weight of objects
two multiplied by the gravitational constant divided by the space between the
two masses squared.
Answer:
The distance of the first bright fringe is given as 
The distance of the second dark fringe from the central bright fringe is given as 
Explanation:
From the question we are told that
The slit separation distance is 
The distance of the slit from the screen is 
The wavelength is 
For constructive interference to occur the distance between the two slit is mathematically represented as

Where m is the order of the fringe which has a value of 1 for first bright fringe
Substituting values


For destructive interference to occur the distance between the two slit is mathematically represented as
![Y_D = [n + \frac{1}{2} ] \frac{\lambda D}{d}](https://tex.z-dn.net/?f=Y_D%20%20%3D%20%20%5Bn%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5D%20%5Cfrac%7B%5Clambda%20D%7D%7Bd%7D)
m = 2
so the formula to get the dark fringe is 
Now substituting values
![Y_D = [ 1 + \frac{1}{2} ] * \frac{633 *10^{-9} * 3.23 }{0.00115}](https://tex.z-dn.net/?f=Y_D%20%3D%20%5B%201%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5D%20%2A%20%5Cfrac%7B633%20%2A10%5E%7B-9%7D%20%2A%203.23%20%7D%7B0.00115%7D)



= Joules ÷ (0.5×Kilograms)
14J ÷ 8.5 = 1.64705882
Remember, 1.64705882 = v², so we need to find the square root.
The square root of 1.64705882 is 1.283377894464448
Hope this helps!