The basic unit that represents the quantum nature of electricity is the charge of the electron, represented with the symbols

, which corresponds to

The quantum nature of electricity was demonstrated for the first time by Millikan, in its oil drop experiment. In this experiment, Millikan put charged oil drops between two metallic plates, applying a potential difference across them, such that the electrical force acting on the drops was in balance with their weight. Knowing the intensity of the electric field and the mass of the drops, he was able to determine the charge of the oil drops, and he found that this charge was always an integer multiple of a certain value, exactly

.
Answer:
Angle of diffraction for second order maxima is θ = 18.941°
Explanation:
From the question it is given that
wavelength of incident light = λ = 541 nm = 541 x 
order of maxima = n =2
diffraction grating has 600 lines per mm
⇒ distance between two slit is
= 1.66 x
m
using the relation of Braggs diffraction formula i.e.,
2dsinθ = nλ ..................................(1)
where, d = distance between two lines of grating
θ is the angle of diffraction
n= order of maxima
λ is the intensity of incident photon
on substituting the respected values in relation (1) we get,
2 x 1.66 x
m sinθ = 2 x 541 x 
⇒ sinθ = 0.3246
⇒ θ =
= 18.941 °
It will go to the bottom of the liquid very slowly. Good Luck!
Answer:
The value is 
Explanation:
From the question we are told that
The width of the slit is 
The distance of the screen from the slit is D = 1.25 m
The width of the central maximum is 
Generally the width of the central maximum is mathematically represented as

Here m is the order of the fringe and given that we are considering the central maximum, the order will be m = 1 because the with of the central maximum separate's the and first maxima
So

=> 
=> 
=> 
Answer:
The answer is given below
Explanation:
Things provided in the statement:
Pressure <em>P1</em> = 120 kPa and <em>P2</em> = 5.6 MP or 5600 kPa
Power, <em>W</em> = 7 kW
Elevation difference = ∆z = 10 m
Mass of flow = m˙
So potential energy changes may be significant
Specific volume of water V= 0.001 m³/kg
Now putting the values in the formula
Power, <em>W </em>= m˙ x V (<em>P1 - P2</em>) + m˙ x g x ∆z
7 = m˙ x 0.001 (5600 - 120 ) + m˙ x 9.8 x 10 x (1 kJ/kg/ 1000 m^2/s^2)
7 = m˙ x 5.48 + m˙ x 0.098
7 = m ˙x 5.38
m˙ = 7/5.38
So mass flow m˙ = 1.301 kJ/s