Answer: 
Explanation:
Given
Cross-sectional area of wire 
Extension of wire 
Extension in a wire is given by

where, 

for same force, length and material

Divide (i) and (ii)

Answer:
Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?
Using Coulomb's law:
F = 1/(4πE) x Q1 x Q2/ r^2
Where
k= 1/(4πE) = 9 x 10^9Nm2/C2
Therefore,
F = 9x 10^9 x 2.50 x 10^-5 x2.50 x
10^-5/. ( 0.5)^2
F= 5.625/ 0.25
F= 22.5N approximately
F= 23N.
To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.
Hence To = 23N, attractive. C ans.
Thanks.
Answer:
option B
Explanation:
given,
diameter of the rotating space = 2 Km
Force exerted at the edge of the space = 1 g
force experienced at the half way = ?
As the object is rotating in the circular part
Force is equal to centripetal acceleration.
at the edge
g = ω² r
ω is the angular velocity of the particle
r is the radius.
now, acceleration at the half way
g' = ω² r'



People at the halfway experience g/2
hence, the correct answer is option B
This depends on the direction of the velocity vector to the magnetic field vector. The force is F=q(VxB) ("x" is the cross product.) The max force is when V and B are perpendicular. Then F=qVB = (1.602e-19)(2000)(300) = 9.612e-14 N