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:
Which of the following is the best example of a good hypothesis?
B. A cheetah can run faster than a tiger This Is A good Hypothesis Because You Can Test this With a Experiment
xXxAnimexXx
Have a great day!
The answer is B artificial selection
Answer:

Explanation:
Firstly, when you measure the voltage across the battery, you get the emf,
E = 13.0 V
In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.
Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (
). But this is not the case.
Let the resistance of the ammeter be r
Hence, using Ohm's law we get the following 2 equations:
.......(1)
......(2)
Substituting the value of r from (2) in (1), we have,

which simplifying gives us,
(which is our required solution)
putting the value of z in either (1) or (2) gives us, r = 0.5325 
Answer:
energy required=-energy lost
energy lost=change in kinetic energy
EL=1/2 mv^2