Answer:
a) 9.72 mm
b) 4.86 mm
Explanation:
wave length of light λ is 580 nm = 580 \times 10⁻⁹ m
Width of slit d = 0.215\times 10⁻³ m
Distance of screen D = 1.8 m.
Width of one fringe = 
Putting the values we get fringe width
= 
=4.86 mm.
a) Width of central maxima = 2 times width of one fringe
= 2 times 4.86
=9.72 mm
b) width of each fringe except central fringe is same , no matter what the order is.Only brightness changes .
So width of either of the two first order bright fringe will be same and it will be
= 4.86 mm.
Answer:
Change in electric potential energy is -28.0 J
Explanation:
Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.
OR
Electric potential energy is also equal to the change in the configuration of the charge particles.
Thus,
Change in electric potential energy = - Work Done
According to the problem, Work Done is equal to 28 J. Thus,
Change in electric potential energy = -28 J
D.all of the above is the answer for this question
The answer is 2 because a and b are slowing it down by condensing it and then freezing it so its not one, b and c are opposites because b slows it down whereas c speeds it up so its not 3, and number 4 is the same explanation for number 3
hope you pass :)
Answer:
Increases
Explanation:
The expression for the capacitance is as follows as;
Here, C is the capacitance, 
is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.
It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.
The expression for the energy stored in the parallel plate capacitor is as follows;
Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.
Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.
Therefore, the option (c) is correct.
