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C. Energy contained in the nucleus of an atom
Explanation:
It is given that,
Frequency of monochromatic light, 
Separation between slits, 
(a) The condition for maxima is given by :

For third maxima,



(b) For second dark fringe, n = 2





Hence, this is the required solution.
Answer:
48.7 J
Explanation:
For a mass-spring system, there is a continuous conversion of energy between elastic potential energy and kinetic energy.
In particular:
- The elastic potential energy is maximum when the system is at its maximum displacement
- The kinetic energy is maximum when the system passes through the equilibrium position
Therefore, the maximum kinetic energy of the system is given by:

where
m is the mass
v is the speed at equilibrium position
In this problem:
m = 3.6 kg
v = 5.2 m/s
Therefore, the maximum kinetic energy is:

The ration of the rms speed of 235uf6 to that of 238uf6 is 1.004.
The molecular mass of 235uf6 is 349, while that of 238uf6 is 352.
The rms speed is calculated as
v=√(3RT/m)
Thus the ratio rms speed of 235uf6 to 238uf6 is calculated as
r=√(352/349)=1.004
Answer:
The value of A is 1.5m/s^2 and B is 0.5m/s^³
Explanation:
The mass of the rocket = 2540 kg.
Given velocity, v(t)=At + Bt^2
Given t =0
a= 1.50 m/s^2
Now, velocity V(t) = A*t + B*t²
If, V(0) = 0, V(1) = 2
a(t) = dV/dt = A+2B × t
a(0) = 1.5m/s^²
1.5m/s^² = A + 2B × 0
A = 1.5m/s^2
now,
V(1) = 2 = A× 1 + B× 1^²
1.5× 1 +B× 1 = 2m/s
B = 2-1.5
B = 0.5m/s^³
Now Check V(t) = A× t + B × t^²
So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s
Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)