Answer:
The average force on ball by the golf club is 340 N.
Explanation:
Given that,
Mass of the golf ball, m = 0.03 kg
Initial speed of the ball, u = 0
Final speed of the ball, v = 34 m/s
Time of contact, 
We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

So, the average force on ball by the golf club is 340 N.
Answer:
a . 0.35cm
b. 11.33cm
Explanation:
a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x
#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

Hence, for currents in same direction, the point is 0.35cm
b. Given both currents flow in opposite directions, the null point lies on the other side.
#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:
Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

Hence, if currents are in opposite directions the point on x-axis is 11.33cm
I would look this one up on Google
Answer:
Explanation:
Given
Force of repulsion between two charge particle is given by force F
Electrostatic force is given by

where
and
is the charges of particle
r=distance between charge particle
when charges are doubled and distance is reduced to half
i.e. q become 2 q and r becomes 0.5 r



Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
q/r
2
, so
E=
(5.11×10
−11
m)
2
(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)
=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.