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.
Answer:
ΔVab = Ed
ΔVab = Va-Vb = Va-V0 = Va
E = Va/ d
= 413V / 0.0795 m
= 5194.97 V/M
Explanation:
the potential difference between two uniform plates is calculated by the formula of electric field.
Single replacement would be represented by a single element being replaced.
This is shown in answer choice B
Where the positions of A and B are swapped
Accuracy?
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Answer: distance d = 4.73e10m
Explanation: Suppose the charge on the black hole is 5740 C which is a positive charge.
Using electric potential V formula:
V = kq / d
Where K = 9.05×10^9Nm^2/C
And e = 1.6×10^-19C
But you don't need to substitute it.
1090 V = 8.99e9N·m²/C² * 5740C /d
Make d the subject of formula
d = 4.73e10 m