Answer:
0.426 volts
Explanation:
It is given that,
The radius of a circular loop, r = 11.2 cm = 0.112 m
An elastic conducting material is stretched into a circular loop.
It is placed with its plane perpendicular to a uniform 0.880 T magnetic field.
The radius of the loop starts to shrink at an instantaneous rate of 68.8 cm/s, dr/dt = 0.688 m/s
We need to find the emf induced in the loop at that instant.
So, the magnitude of induced emf is 0.426 volts.
Picture? I may be able to answer if you have a chart or some kind of graph as a referral to the question
The amplitude of a wave corresponds to its maximum oscillation of the wave itself.
In our problem, the equation of the wave is
![y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])](https://tex.z-dn.net/?f=y%28x%2Ct%29%3D%20%280.750cm%29cos%28%5Cpi%20%5B%280.400cm-1%29x%2B%28250s-1%29t%5D%29)
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
and therefore this value corresponds to the amplitude of the wave.
This question requires the use of the equation of motion:
v = u + at [v is final velocity (0), u is initial velocity (24), a is acceleration, t is time (13)]
to calculate the acceleration. This can then be multiplied by the mass of the plane to obtain the net force via:
F = ma (F is force, m is mass, a is acceleration)
First, we calculate the acceleration:
0 = 24 + 13(a)
a = -24/13 m/s^2
The force is then:
F = 90000 * (-24/13)
F = -1.66*10^5 Newtons
The negative sign indicates that the force and acceleration are in the opposite direction as the velocity (since we took velocity to be positive)
