1. F = 8 Hz
2. V = 8 m/s
3. D = 8 m
4. T = 8 sec
(1) Doubling of the current through the wire will result in doubling of its magnetic field.
The magnetic field around a wire is a function of the current I and radial distance r

(with mu denoting the magnetic permeability of the medium). So, B is directly proportional to I. The field magnitude will double with the doubled current from 5A to 10A
(2) Using the same formula as in (1), we can see that the magnetic field is inversely proportional to the radial distance from the wire. So, a particle at 20cm will experience half the magnitude compared to a particle at 10cm.
(3) Answer
If a particle with a charge q moves through a magnetic field B with velocity v, it will be acted on by the magnetic force

So, a particle with charge -2uC will experience a magnetic force of same magnitude but opposite direction (and perpendicular to B) as compared to a particle with a charge of 2uC
If<span> a neutral </span>object loses<span> some </span>electrons<span>, </span>then<span> it will possess more protons</span>
Answer: -0.84 rad/sec (clockwise)
Explanation:
Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:
L1 = L2
L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m
L1 = -521.15 kg.m2/sec (1)
(Considering to the man as a particle that is moving opposite to the rotation of the turntable, so the sign is negative).
Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:
Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2
The total angular momentum, once the runner has come to an stop, can be written as follows:
L2= (It + Im) ωf = -521.15 kg.m2/sec
L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec
Solving for ωf, we get:
ωf = -0.84 rad/sec (clockwise)
Answer:
B. Moon
Explanation:
In physics, that would be the answer.