The total quantity of electrons that have flowed through a circuit is a
quantity of charge, measured in Coulombs, or in Ampere-seconds.
The <em><u>rate</u></em> of flow of electrons, or more accurately the rate of flow of
the charge on them, is electrical current. Its unit is the Ampere.
1 Ampere is 1 Coulomb of charge per second.
The first rubber balloon was made by Professor Michael Faraday in 1824, out of two sheets of rubber whose edges were pressed together. Hot air balloonwas the balloon to make the first recorded manned flight. It was made by the Montgolfier brothers and launched on 21 November 1783.
The energy of a photon is given by

where

is the Planck constant
f is the frequency of the photon
In our problem, the frequency of the light is

therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
<h2>
Hey There!</h2><h2>
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Answer:</h2><h2 /><h2>

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<h2>DATA:</h2>
mass = m = 2kg
Distance = x = 6m
Force = 30N
TO FIND:
Work = W = ?
Velocity = V = ?
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SOLUTION:</h2>
According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.
To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

Base is the x-axis of the graph which is Position i.e. 6m
Height is the y-axis of the graph which is Force i.e. 30N
So,

W = 90 J
The work done is 90 J.
According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.



<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>