Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of v=0.8c
1 answer:
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Answer:
The magnitude of the second charge is or
Explanation:
The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.
The electric potential energy of some charge at a point in the electric field of another charge
is given by the product of the amount of charge
and electric potential at that point due to the charge
.
The electric potential at that point is given by
where is the Coulomb's constant.
Therefore,
Now, We have given two charges and
, whose value is to be found.
When the two charges are infinitely dar apart, the electric potential energy of the system is given by
When the coordinates of position of the two charges are
The distance between the two charges is given by
The electric potential energy of the charges in this configuration is given by
The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.
Therefore,
Answer
given,
Mass of the runner, M = 70 Kg
speed of the runner on the second base = 4.35 m/s
speed at the base = 0 m/s
Acceleration due to gravity,g = 9.8 m/s²
a) magnitude of mechanical energy lost
Mechanical energy lost is equal top gain in kinetic energy
b) Work done = Force x displacement
W = F. x
F = μ mg
W = μ mg . x
Work done is equal to 662.29 J
using the coefficient of the friction,μ = 0.7
x = 1.38 m
Hence, the runner will slide to 1.38 m.