Answer:
(b) 10 Wb
Explanation:
Given;
angle of inclination of magnetic field, θ = 30°
initial area of the plane, A₁ = 1 m²
initial magnetic flux through the plane, Φ₁ = 5.0 Wb
Magnetic flux is given as;
Φ = BACosθ
where;
B is the strength of magnetic field
A is the area of the plane
θ is the angle of inclination
Φ₁ = BA₁Cosθ
5 = B(1 x cos30)
B = 5/(cos30)
B = 5.7735 T
Now calculate the magnetic flux through a 2.0 m² portion of the same plane
Φ₂ = BA₂Cosθ
Φ₂ = 5.7735 x 2 x cos30
Φ₂ = 10 Wb
Therefore, the magnetic flux through a 2.0 m² portion of the same plane is is 10 Wb.
Option "b"
Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.
Rate of speed that toy is moving is irelevant.
childs force is:
Fc = 2N
Fc = Ff (Ff -friction force)
Ff = a*Q
where Q is weight of the toy and a is friction
if we express a we get
a = F/Q = 2/8 = 0.25
Answer:
<em>radius of the loop = 7.9 mm</em>
<em>number of turns N ≅ 399 turns</em>
Explanation:
length of wire L= 2 m
field strength B = 3 mT = 0.003 T
current I = 12 A
recall that field strength B = μnI
where n is the turn per unit length
vacuum permeability μ = 
= 1.256 x 10^-6 T-m/A
imputing values, we have
0.003 = 1.256 x 10^−6 x n x 12
0.003 = 1.507 x 10^-5 x n
n = 199.07 turns per unit length
for a length of 2 m,
number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>
since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.
this length is also equal to the circumference of each loop
the circumference of each loop = 
0.005 = 2 x 3.142 x r
r = 0.005/6.284 = 
= 0.0079 m =<em> 7.9 mm</em>
Well this question looks like it makes some assumptions. So assuming that both cars have the same mass and experience the same wind resistance regardless of speed and same internal frictions, then we could say "The car that finishes last has the lowest power". The reason is that for a given race the cars must overcome losses associated with motion. Since they all travel the same distance, the amount of work will be the same for both. This is because work is force times distance. If the force applied is the same in both cases (identical cars with constant wind resistance) and the distance is the same for both (a fair race track) then W=F·d will be the same.
Power, however, is the work done divided by the time over which it is done. So for a slower car, time t will be larger. The power ratio W/t will be smaller for the longer time (slower car).
I would say that the answer to this is the last option: ALL OF THE ABOVE. A lean fuel or air mixture, incorrect spark plug heat range, and a blocked carburetor bowl vent would be among the reasons of an engine to surge and hunt at the top no-load speeds. Hope this answer helps.
