The stage where atoms are spread out and bouncy is the gas stage.
a light wave travels from air into a medium of refractive index 1.54. If the wavelength of the light in air is 5.4 * 10^-7. calc
1 answer:
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Answer:
Electric field at radius r inside the solid sphere is
Electric field at radius r between inner radius and outer radius of the shell is
E=0 N/C
Explanation:Given
The radius of the solid sphere is
The charge on the solid sphere is
The inner radius of the shell is
The outer radius of the shell is
The total charge on the shell is
PART(A)
The magnitude of electric field at radius r where \\The volumetric charge density of the solid sphere will be
The charge enclosed by the radius r inside the solid sphere is
rho=
According to gauss law
PART(B)
The electric field at radius r where
The shell is conducting so due to induction of charge
The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell
So,
The charge on the inner surface of the shell is
Due to conservation of the charge on the shell
The charge accumulated on the outer surface of the shell is
The charge enclosed by the radius r where
According to gauss law
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Answer:
Explanation:
Since work done is in the form of potential energy, we will use the formula of potential energy here.
We know that,
<h3>P.E. = mgh </h3>Where,
m = mass = 20 kg
g = acceleration due to gravity = 10 m/s²
h = vertical height = 20 m
So,
<h3>Work done = mgh</h3>Work done = (20)(10)(20)
Work done = 4000 joules
Work done = 4 kJ