Answer:
-611.32 N/C
0.43723 m
Explanation:
k = Coulomb constant = 
q = Charge = -4.25 nC
r = Distance from particle = 0.25 m
Electric field is given by
The magnitude is 611.32 N/C
The electric field will point straight down as the sign is negative towards the particle.
The distance from the electric field is 1.71436 m
<h2>
Speed with which it return to its initial level is 100 m/s</h2>
Explanation:
We have equation of motion v² = u² + 2as
Initial velocity, u = 100 m/s
Acceleration, a = -9.81 m/s²
Final velocity, v = ?
Displacement, s = 0 m
Substituting
v² = u² + 2as
v² = 100² + 2 x -9.81 x 0
v² = 100²
v = ±100 m/s
+100 m/s is initial velocity and -100 m/s is final velocity.
Speed with which it return to its initial level is 100 m/s
Answer:
The momentum before is equal to the momentum after
Explanation:
It is equal and should level out in an equation.
Answer:
the needle will direct its North South according to the magnetic field of current carrying wire.
Explanation:
A current carrying wire always has a magnetic field around it, in circular loops. This magnetic field will be either clockwise or anticlockwise depending on the direction of current.
Right hand rule tells the direction. Place the current carrying wire in your right hand with thumb pointing the direction of current. Curl of the fingers tell the direction of current.
When the needle gets in the vicinity of the field, its poles aligns itself with the field. (previous position of the compass needle has no effect on its position in the field). The north pole and south pole will be set in the direction of magnetic field.
The distance between the needle and wire does effect the strength (accuracy) of the needle position. Strong field will create strong deflection of the needle whereas when the distance from wire increases, field weakens, thus the deflection of needle will be weak.
