Answer:
Correct, is there another part to the question?
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:
170 W
Explanation:
Applying
P = VI.................... Equation 1
Where P = Power generated in watt, V = Voltage supplied to the circuit, I = Current running through the circuit.
From the question,
Given: V = 17 V, I = 10 A
Substitute these values into equation 1
P = (17×10)
P = 170 Watt.
Hence the power generated is 170 W.
The right option is A. 170 W
Answer: 5.5m/s
Explanation:
vf=vi+at
vf= 4.0m/s + (0.50m/s^2)(3.0s)