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:
y = 3.16 m
Explanation:
Once thrown straight up, the only influence on the ball is gravity, which acts slowing down the ball.
In absence of gravity, the ball had moved forever with the same initial velocity, so the displacement at a time t had been as follows:
y = v₀*t
Due to the action of gravity, the actual displacement is less than this value, and can be calculated with the following equation:
Replacing by v₀ = 12. 0 m/s, t = 0.300 s, and g = 9.8 m/s², we can solve for the vertical displacement, as follows:
So, the position (taking the origin coincident with the point where the ball leaves the hand), is just the vertical displacement we have just found.
⇒ y = 3.16 m
Answer:
Explanation:
Volume ,V = 3 L
Initial temperature ,T₁ = 273 K
Initial pressure ,P₁ = 105 kPa
Final temperature ,T₂ = 381 K
Lets take final pressure =P₂
We know that ,If the volume of the gas is constant ,then we can say that
Now by putting the values in the above equation we get
Therefore the final pressure will be 146.53 kPa.
Answer:19.997m/s
Explanation:
Velocity=√(centripetal acceleration×radius)
Velocity=√(13.33×30)
Velocity=√(399.9)
Velocity=19.997