Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a).
= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now,
A' = amplitude = 1.4142 m
b).
m' = 2m
Hence,
c).
Therefore, factor
Thus, the energy will change half times as the result of the collision.
Answer:
Option c) are perpendicular to the electric field
Explanation:
Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90 to the equipotential surface.
Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.
Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.
Answer with Explanation:
a. Option d is true.
a negatively charged plane parallel to the end faces of the cylinder
b. Radius of cylinder, r=0.66m
Magnitude of electric field, E=300 N/C
We have to find the net flux through the closed surface.
Net electric flux,
c.
Net charge,
Where
Where