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: a) the greater speed for the ball is getting with the large radius of the circle. b) 1.68* 10 ^3 m/s^2 c) 1.25*10^3 m/s^2
Explanation: In order to solve this problem firstly we have to consider that speed in a of the circular movement is directly the angular rotation multiply the radius of the circle so by this we found that the second radius get large speed.
Secondly to calculate the centripetal acceleration for the ball we have to considerer the relationship given by:
acceleration in a circular movement= ω^2*r
so
a1= (8.44 *2*π)^2*r1=1.68 *10^3 m/s^2
a2= (5.95*2*π)^2*r2=1.25*10^3 m/s^2
Answer:
A
Explanation:
if he goes to the west, the east is opposite so 100-30
Let t = Theta and p = Phi
Tan t = y/x Then x =y/Tant.
Tant = y/(x-d) x-d = y/Tanp
y/Tant - d = y/Tanp
y -d*Tanr = y*Tant/Tanp
y-y*Tant/Tanp = d*Tanr
y(1 - Tanr/Tanp = d*Tant
y = d*Tant/(1-Tant/Tanp)