Answer:
X: Always attractive
Y: Infinite range
Z: Attractive or repulsive
ANSWER IS C
The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to

where

is the angle between the directions of v and B.
1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is

. In this case, v and B are perpendicular, so

, therefore we have:

2) In this second case, the angle between v and B is

. The charge is now

, and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
Answer:
The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Explanation:
Given;
first magnetic field at first distance, B₁ = 2.50 mT
first distance, r₁ = 12.6 cm = 0.126 m
Second magnetic field at Second distance, B₂ = ?
Second distance, r₂ = ?
Magnetic field for a straight wire is given as;

Where:
μ is permeability
B is magnetic field
I is current flowing in the wire
r distance to the wire

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Friction is caused by the uneven surfaces of touching objects
When Janet leaves the platform, she's moving horizontally at 1.92 m/s. We assume that there's no air resistance, and gravity has no effect on horizontal motion. There's no horizontal force acting on Janet to make her move horizontally any faster or slower than 1.92 m/s.
She's in the air for 1.1 second before she hits the water.
Moving horizontally at 1.92 m/s for 1.1 second, she sails out away from the platform
(1.92 m/s) x (1.1 sec) = <em>2.112 meters</em>