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)
Answer:

Explanation:
In this question we have given

we have to find

We know that
optical path difference for bright fringe is given as
Here,
n is order of fringe
and optical path difference for dark fringe is given as
since the light with wavelength
produces its third-order bright fringe at the same place where the light with wavelength
produces its fourth dark fringe
it means
optical path difference for 3rd order bright fringe= optical path difference for forth order dark fringe
Therefore,
...............(1)
Put value of
in equation (1)



Given mass = 2kg, height = 10m,g = 9.8.
We know that Work done W = FD
= > W = (mg)(D)
= > W = (2 * 9.8)(10)
= > W = 196 Joules.
Hope this helps!
Answer:
v= 3.18 m/s
Explanation:
Given that
m= 150 g = 0.15 kg
M= 240 g = 0.24 kg
Angular speed ,ω = 150 rpm
The speed in rad/s


ω = 15.7 rad/s
The distance of center of mass from 150 g

r= 20.30 cm
The speed of the mass 150 g
v= ω r
v= 20.30 x 15.7 cm/s
v= 318.71 cm/s
v= 3.18 m/s
Can you please post the picture of the graph and i will gladly answers it.