It’s the engery. you are changing the engery
This interaction is known as <em>constructive interference</em>. It's a result of linear superposition.
The answer is : C ) air,water,and the tank glass.
-Hope this helps.
Part a)
At t = 0 the position of the object is given as
![x = 0](https://tex.z-dn.net/?f=%20x%20%3D%200)
At t = 2
![x = 2 sin(\pi/2) = 2cm](https://tex.z-dn.net/?f=%20x%20%3D%202%20sin%28%5Cpi%2F2%29%20%3D%202cm)
so displacement of the object is given as
![d = 2 - 0 = 2cm](https://tex.z-dn.net/?f=d%20%3D%202%20-%200%20%3D%202cm)
so average speed is given as
![v_{avg} = \frac{2}{2} = 1 cm/s](https://tex.z-dn.net/?f=v_%7Bavg%7D%20%3D%20%5Cfrac%7B2%7D%7B2%7D%20%3D%201%20cm%2Fs)
Part b)
instantaneous speed is given by
![v = \frac{dy}{dt}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7Bdy%7D%7Bdt%7D)
![v = 2cos(\pi t/4 ) * \frac{\pi}{4}](https://tex.z-dn.net/?f=v%20%3D%202cos%28%5Cpi%20t%2F4%20%29%20%2A%20%5Cfrac%7B%5Cpi%7D%7B4%7D)
now at t= 0
![v = \frac{\pi}{2} cm/s](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20cm%2Fs)
at t = 1
![v = 2 cos(\pi/4) * \frac{\pi}{4}](https://tex.z-dn.net/?f=%20v%20%3D%202%20cos%28%5Cpi%2F4%29%20%2A%20%5Cfrac%7B%5Cpi%7D%7B4%7D)
![v = \frac{\pi}{2\sqrt2}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%5Csqrt2%7D)
at t = 2
![v = 0](https://tex.z-dn.net/?f=v%20%3D%200)
Part c)
Average acceleration is given as
![a_{avg} = \frac{v_f - v_i}{t}](https://tex.z-dn.net/?f=a_%7Bavg%7D%20%3D%20%5Cfrac%7Bv_f%20-%20v_i%7D%7Bt%7D)
![a_{avg} = \frac{0 - \frac{\pi}{2}}{2}](https://tex.z-dn.net/?f=a_%7Bavg%7D%20%3D%20%5Cfrac%7B0%20-%20%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7B2%7D)
![a = -\frac{\pi}{4} cm/s^2](https://tex.z-dn.net/?f=a%20%3D%20-%5Cfrac%7B%5Cpi%7D%7B4%7D%20cm%2Fs%5E2)
Part d)
Now for instantaneous acceleration
As we know that
![a =- \omega^2 y](https://tex.z-dn.net/?f=a%20%3D-%20%5Comega%5E2%20y)
at t = 0
![a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2](https://tex.z-dn.net/?f=%20a%20%3D%20-%5Cfrac%7B%5Cpi%5E2%7D%7B16%7D%20%2A%200%20%3D%200%20cm%2Fs%5E2)
at t = 1
![y = \sqrt2 cm](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt2%20cm)
now we have
![a = -\frac{\pi^2}{16}*\sqrt2](https://tex.z-dn.net/?f=a%20%3D%20-%5Cfrac%7B%5Cpi%5E2%7D%7B16%7D%2A%5Csqrt2)
At t = 2 we have
![y = 2 cm](https://tex.z-dn.net/?f=%20y%20%3D%202%20cm)
![a = -\frac{\pi^2}{16}*2](https://tex.z-dn.net/?f=a%20%3D%20-%5Cfrac%7B%5Cpi%5E2%7D%7B16%7D%2A2)
![a = -\frac{\pi^2}{8}](https://tex.z-dn.net/?f=a%20%3D%20-%5Cfrac%7B%5Cpi%5E2%7D%7B8%7D)
<em>so above is the instantaneous accelerations</em>
The concept that we need here to give a proper solution is mutual inductance.
The mutual inductance is given by the expression
![M=\frac{N\Phi}{I}](https://tex.z-dn.net/?f=M%3D%5Cfrac%7BN%5CPhi%7D%7BI%7D)
Where,
I = current
N = Number of turns
Flux through the solenoid.
Part A) Then we have in our values that,
![I=6.6A](https://tex.z-dn.net/?f=I%3D6.6A)
![\Phi= 3.50*10^{-2}Wb](https://tex.z-dn.net/?f=%5CPhi%3D%203.50%2A10%5E%7B-2%7DWb)
![N=450](https://tex.z-dn.net/?f=N%3D450)
Replacing in the equation,
![M = \frac{450*350*10^{-2}}{6.60}](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7B450%2A350%2A10%5E%7B-2%7D%7D%7B6.60%7D)
![M = 2.39H](https://tex.z-dn.net/?f=M%20%3D%202.39H)
Part B) Here is required the Flux, then using the same expression we have that
![\Phi = \frac{IM'}{N}](https://tex.z-dn.net/?f=%5CPhi%20%3D%20%5Cfrac%7BIM%27%7D%7BN%7D)
We conserve the same value for the Inductance but now we have a current of 2.6, then
![\Phi = \frac{2.6*2.39}{690}](https://tex.z-dn.net/?f=%5CPhi%20%3D%20%5Cfrac%7B2.6%2A2.39%7D%7B690%7D)
![\Phi = 9*10^{-3}Wb](https://tex.z-dn.net/?f=%5CPhi%20%3D%209%2A10%5E%7B-3%7DWb)
Therefore the flux in Solenoid 1 is ![9*10^{-3}Wb](https://tex.z-dn.net/?f=9%2A10%5E%7B-3%7DWb)