Answer:
The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Explanation:
Given that,
Wavelength = λ
For D to be small,
We need to calculate the minimum width
Using formula of minimum width


Where, D = width of slit
= wavelength
Put the value into the formula

Here,
should be maximum.
So. maximum value of
is 1
Put the value into the formula


(b). If the minimum number is 50
Then, the width is


(c). If the minimum number is 1000
Then, the width is


Hence, The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Answer:you in connections too?
Explanation:
Answer:
0.80 m
Explanation:
elastic potential energy formula
elastic potential energy = 0.5 × spring constant × (extension) 2
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (0.314 kg) x (164 m/s²)
= 51.5 newtons
(about 11.6 pounds).
Notice that the ball is only accelerating while it's in contact with the racket. The instant the ball loses contact with the racket, it stops accelerating, and sails off in a straight line at whatever speed it had when it left the strings.
~ I hope this helped, and I would appreciate Brainliest. ♡ ~ ( I request this to all the lengthy answers I give ! )