Answer:
Change in potential energy = 7350 Joules
Explanation:
It is given that,
Side of cube, a = 0.5 m
Density of cube, 
The cube is lifted vertically by a crane to a height of 3 m
We know that, density 
So, m = d × V (V = volume of cube = a³)
m = 250 kg
We have to find the change in potential energy of the cube. At ground level, the potential energy is equal to 0.
Potential energy at height h is given by :
PE = 250 kg × 9.8 m/s² ×3 m
PE = 7350 Joules
So, change in potential energy of the cube is 7350 Joules.
Acceleration = velocity/time
A= 3.5m/s/15s
A= 0.23m/s^2
60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg
Now,new weight with g = 6m/s^2
=m×g' (here g' is new acceleration of the new planet)
= 10×6=60N
Answer:
Explanation:
1 )
Here
wave length used that is λ = 580 nm
=580 x 10⁻⁹
distance between slit d = .46 mm
= .46 x 10⁻³
Angular position of first order interference maxima
= λ / d radian
= 580 x 10⁻⁹ / .46 x 10⁻³
= 0.126 x 10⁻² radian
2 )
Angular position of second order interference maxima
2 x 0.126 x 10⁻² radian
= 0.252 x 10⁻² radian
3 )
For intensity distribution the formula is
I = I₀ cos²δ/2 ( δ is phase difference of two lights.
For angular position of θ1
δ = .126 x 10⁻² radian
I = I₀ cos².126x 10⁻²/2
= I₀ X .998
For angular position of θ2
I = I₀ cos².126x2x 10⁻²/2
= I₀ cos².126x 10⁻²
Answer:
Explanation:
The momentum change is defined as:
Taking the downward motion as negative and the upward motion as positive, we have:
Replacing (2) and (3) in (1):
