Answer:
Explanation:
The expression which represent the first diffraction minima by a circular aperture is given by 
--------eqn 1
The angle through which the first minima is diffracted is given by 
---------eqn 2
As 
is very small so we can write 
So from eqn 1 and eqn 2 we can write
--------eqn 3
Here 
is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture
It is given that diameter of circular aperture is 14.7 cm so 
Now putting all these value in eqn 3
<h3>
</h3><h3>Given</h3>
v = 20m\s
a = 3m\s^2
t = 4sec
Firstly we have to find u
a = 
3m\s =
12m\s = 20 - u
20 - u = 12m\s
- u = -8
u = 8
Now we can easily find distance by using second equation of motion
s = ut + 1\2 at^2
s = 8(4) + 1\2(3)(16)
s = 32 + 24
s = 56
So distance is 56 m\s hope it helps
The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.
C = Q/ΔV
C is the capacitance
Q is the stored charge
ΔV is the potential difference
Rearrange the equation:
ΔV = Q/C
We also know the capacitance of a parallel-plate capacitor is given by:
C = κε₀A/d
C is the capacitance
κ is the capacitor's dielectric constant
ε₀ is the electric constant
A is the area of the plates
d is the plate separation
If we substitute C:
ΔV = Qd/(κε₀A)
We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.
