Answer:
<em>c. ABBA counterbalancing
</em>
Explanation:
The student should not use the method because it is a progressive error management technique for each subject by introducing all <em>treatment circumstances twice, first in one sequence, then in the other (AB, BA) by subject counterbalancing.</em>
If participants experience conditions more than once, they experience the conditions first in one order, then the opposite order.
Theoretically, 35 x 18 = 630
1). The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.
2). The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.
3). The block is also affected by friction with the air (air resistance) as it
falls to the ground. Friction robs some kinetic energy.
Answer:
(a) 1.58 V
(b) 0.0126 Wb
(c) 0.0493 V
Solution:
As per the question:
No. of turns in the coil, N = 400 turns
Self Inductance of the coil, L = 7.50 mH = 
Current in the coil, i = ![1680cos[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=1680cos%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
A
where
Now,
(a) To calculate the maximum emf:
We know that maximum emf induced in the coil is given by:
![e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20L%5Cfrac%7Bd%7D%7Bdt%7D%281680%29cos%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
![e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20-%207.50%5Ctimes%2010%5E%7B-%203%7D%5Ctimes%20%5Cfrac%7B%5Cpi%7D%7B0.0250%7D%5Ctimes%20%5Cfrac%7Bd%7D%7Bdt%7D%281680%29sin%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
For maximum emf, 
should be maximum, i.e., 1
Now, the magnitude of the maximum emf is given by:
(b) To calculate the maximum average flux,we know that:
(c) To calculate the magnitude of the induced emf at t = 0.0180 s:
