Answer:
The force exerted on the
is 
Explanation:
From the question we are told that
The area is 
The magnitude of charge placed on them is 
The charge placed between the plate is 
The electric field generated around the plate is mathematically represented as

Substituting values


The force exerted the charge
is mathematically represented as

Substituting values


Answer:
while ice is made by water again it melts and becomes water. water is colourless and odourless and has no taste but ice is only cold and hard. water is used for drinking and other things. but is for freshness and it never flows
Explanation:
so ice reflect more energy compared to water
To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.
Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

The area of a sphere is given by

So replacing we have to

Since the question tells us to find the proportion when

So considering the two intensities we have to


The ratio between the two intensities would be

The power does not change therefore it remains constant, which allows summarizing the expression to

Re-arrange to find 



Therefore the intensity at five times this distance from the source is 
A transformer is used to increase or decrease a voltage. BUT ... it has to be an AC voltage, or the transformer doesn't work.