(a) The net flux through the coil is zero.
In fact, the magnetic field generated by the wire forms concentric circles around the wire. The wire is placed along the diameter of the coil, so we can imagine as it divides the coil into two emisphere. Therefore, the magnetic field of the wire is perpendicular to the plane of the coil, but the direction of the field is opposite in the two emispheres. Since the two emispheres have same area, then the magnetic fluxes in the two emispheres are equal but opposite in sign, and so they cancel out when summing them together to find the net flux.
(b) If the wire passes through the center of the coil but it is perpendicular to the plane of the wire, the net flux through the coil is still zero.
In fact, the magnetic field generated by the wire forms concentric lines around the wire, so it is parallel to the plane of the coil. But the flux is equal to

where

is the angle between the direction of the magnetic field and the perpendicular to the plane of the coil, so in this case

and so the cosine is zero, therefore the net flux is zero.
I'm going to say false hope that helped
<em>Answer:</em>
<em>Velocity is vector quantity.So it needs direction in addition to speed.</em>
<em>Explanation:</em>
<em>The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of its speed and direction of motion. </em>
Answer:
Explanation:
See the attachment for the details. A right triangle is formed to find the hypotenuse of the two legs consisting of the actual driving distances and times. The hypotenuse gives the vector information for the displacement at the end of 8 hours of driving.
The individual driving times and distances are summed to provide:
(<u>a) How far did he travel?</u>
103 km
<u>(b) What was his average speed?</u>
12.88 km/h
<u>(c) What was his displacement?</u>
73.82 km
<u>(d) What was his average velocity?</u>
9.228 km/h
Answer:
The correct answer is C. 45.5 lbs.
Explanation:
In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.
The formula for any problem involving a lever is:

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.
The parameter of the formula that you need is F_l:

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

F_l=45.5 lbs