Answer:
The acceleration of man 1 and 2 is and
.
Explanation:
Mass of man 1, m₁ = 80 kg
Mass of man 2, m₂ = 60 kg
One man pulls on the rope with a force of 250 N.
Let a₁ is acceleration of man 1,
F = m₁a₁
Let a₂ is acceleration of man 1,
F = m₂a₂
So, the acceleration of man 1 and 2 is and
.
A) 4.7 cm
The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is
where
n is the order of the maximum
is the wavelength
is the distance between the slits
In this problem,
n = 5
So we find
And given the distance of the screen from the slits,
The distance of the 5th bright fringe from the central bright fringe will be given by
B) 8.1 cm
The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:
To find the angle corresponding to the 8th dark fringe, we substitute n=8:
And the distance of the 8th dark fringe from the central bright fringe will be given by