Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
So,
a) 0 < r < r1 :
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
Hence, E = 0 for r < r1
b) r1 < r < r2:
Electric field =?
Let, us consider the Gaussian Surface,
E x 4
= 
So,
Rearranging the above equation to get Electric field, we will get:
E = 
Multiply and divide by
E =
x 
Rearranging the above equation, we will get Electric Field for r1 < r < r2:
E= (σ1 x
) /(
x
)
c) r > r2 :
Electric Field = ?
E x 4
= 
Rearranging the above equation for E:
E = 
E =
+ 
As we know from above, that:
= (σ1 x
) /(
x
)
Then, Similarly,
= (σ2 x
) /(
x
)
So,
E =
+ 
Replacing the above equations to get E:
E = (σ1 x
) /(
x
) + (σ2 x
) /(
x
)
Now, for
d) Under what conditions, E = 0, for r > r2?
For r > r2, E =0 if
σ1 x
= - σ2 x 
Answer:
Explanation:
Atmospheric pressure = 7 x 10⁴ Pa
force on a disk-shaped region 2.00 m in radius at the surface of the ocean due to atmosphere = pressure x area
= 7 x 10⁴ x 3.14 x 2 x 2
= 87.92 x 10⁴ N
b )
weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m
Pressure x area
height x density x acceleration of gravity x π r²
= 10 x 415 x 6.2 x 3.14 x 2 x 2
=323168.8 N
c ) Pressure at a depth of 10 m
atmospheric pressure + pressure due to liquid column
= 7 x 10⁴ + 10 x 415 x 6.2 ( hρg)
= 7 x 10⁴ + 10 x 415 x 6.2
(7 + 2.57 )x 10⁴ Pa
9.57 x 10⁴ Pa
Answer:
deductive reasoning usually follows steps .
- That is, how we predict what the observations should be if the theory were correct