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: from the land to the ocean
Explanation: Apex
Simply, apply the formula
and insert the values of m = mass, v = velocity and E = Energy.
The result will be
, m = 1 kg
<h3><u>Answer;</u></h3>
= 75 %
<h3><u>Explanation</u>;</h3>
To measure efficiency of a machine, compare the work output to work input.
Efficiency = Work output/ work output × 100 %
= 300/400 × 100
= 75 %
Answer:
The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.
Explanation:
Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively
and the mass of big rock be 'M'
Initial momentum of the system equals
Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'
Thus the final momentum of the system is
Equating initial and the final momenta we get
Now since the surface is frictionless thus the energy is also conserved thus
Similarly the final energy becomes
\
Equating initial and final energies we get
Solving i and ii we get
Using this in equation i we get
Thus putting v = -v' in equation i we get V' = 0
This implies Smaller stone rebounds while as larger stone remains stationary.