Answer:
83.2 W/m^2
Explanation:
The radiation per unit area of a star is directly proportional to the power emitted, which is given by Stefan-Boltzmann law:
where
is the Stefan-Boltzmann constant
A is the surface area
T is the surface temperature
So, we see that the radiation per unit area is proportional to the fourth power of the temperature:
So in our problem we can write:
where
is the power per unit area of the present sun
is the temperature of the sun
is the power per unit area of sun X
is the temperature of sun X
Solving for I2, we find
The particles always move parallel and perpendicular to the waves. The waves which are in the water moves a circle. Both up and down and back and forth.
Good luck :)
components of the speed of the coin is given as
now the time taken by the coin to reach the plate is given by
now in order to find the height
so it is placed at 1.52 m height
I'm sure you've noticed that an airplane high in the sky, far away
from you, looks like it's moving very slowly. At the same time,
somebody passing you on a skateboard whizzes past you at
high speed. The farther away something is from you, the slower
it appears to move.
The nearest star outside the solar system is almost 32 thousand times
as far away from us as the farthest visible planet (Saturn) is, and all of the
other stars are farther than that.
That's why you have to wait a few thousand years before you notice
that the shape of a constellation has changed.
To put it a slightly different way . . . Everything is in motion. The motion is
more noticeable for nearby things, and less noticeable for farther-away things.
Objects within our solar system are the only ones near enough so that a human
lifetime is a long enough period in which to notice the change in their position.
Even Pluto moves less then 1.5° against the 'background' stars in a whole year.
This all makes me feel small. How about you ?
Answer:
(a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is 
Explanation:
Given that,
Charge q₁ = -4.00 μC
Inner radius = 3.13 m
Outer radius = 4.13 cm
Net charge q₂ = -6.43 μC
We need to calculate the charge on the outer surface
Using formula of charge
The charge on the inner surface is q.
We need to calculate the electric field outside the shell
Using formula of electric field
Put the value into the formula
Hence, (a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is
