Step-#1:
Ignore the wire on the right.
Find the strength and direction of the magnetic field at P,
caused by the wire on the left, 0.04m away, carrying 5.0A
of current upward.
Write it down.
Step #2:
Now, ignore the wire on the left.
Find the strength and direction of the magnetic field at P,
caused by the wire on the right, 0.04m away, carrying 8.0A
of current downward.
Write it down.
Step #3:
Take the two sets of magnitude and direction that you wrote down
and ADD them.
The total magnetic field at P is the SUM of (the field due to the left wire)
PLUS (the field due to the right wire).
So just calculate them separately, then addum up.
Answer:
The second vector 
points due West with a magnitude of 600N
Explanation:
The original vector 
points with a magnitude of 200N due east, the Resultant vector 
points due west (that's how east/west direction can be interpreted, from east to west) with a magnitude of 400N. If we choose East as the positive direction and West as the negative one, we can write the following vectorial equation:
With the negative sign signifying that the vector points west.
The answer is C.
The Kinetic energy which was exerted and experience pulling the string of a bow is kept as a potential energy at the end of the arrow in contact with the string. Once release from aim at stationary position the potential energy is again transformed.
If it;s a good insulator, there'll be no heat transfer warm to cold. So, over time, given the insulation ... nothing should happen ...
Ahhh this going to be confusing sorry...
1. α = Δω / Δt = 28 rad/s / 19s = 1.47 rad/s²
2. Θ = ½αt² = ½ * 1.47rad/s² * (19s)² = 266 rads
3. I = ½mr² = ½ * 8.7kg * (0.33m)² = 0.47 kg·m²
4. ΔEk = ½Iω² = ½ * 0.47kg·m² * (28rad/s)² = 186 J
5. a = α r = 1.47rad/s² * 0.33m = 0.49 m/s²
6. a = ω² r = (14rad/s)² * 0.33m = 65 m/s²
7. v = ω r = 28rad/s * ½(0.33m) = 4.62 m/s
8. s = Θ r = 266 rads * 0.33m = 88 m
