Answer:
The induced emf is
Explanation:
From the question we are told that
The radius of the circular loop is ![r = 9.50 \ cm = 0.095 \ m](https://tex.z-dn.net/?f=r%20%3D%20%209.50%20%5C%20cm%20%20%3D%20%200.095%20%5C%20m)
The intensity of the wave is ![I = 0.0215 \ W/m^2](https://tex.z-dn.net/?f=I%20%20%3D%20%200.0215%20%5C%20W%2Fm%5E2)
The wavelength is ![\lambda = 6.90\ m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%206.90%5C%20m)
Generally the intensity is mathematically represented as
![I = \frac{ c * B^2 }{ 2 * \mu_o }](https://tex.z-dn.net/?f=I%20%20%3D%20%20%5Cfrac%7B%20c%20%2A%20%20B%5E2%20%20%7D%7B%202%20%2A%20%5Cmu_o%20%20%7D)
Here
is the permeability of free space with value
![\mu_o = 4 \pi *10^{-7} N/A^2](https://tex.z-dn.net/?f=%5Cmu_o%20%20%3D%20%204%20%5Cpi%20%2A10%5E%7B-7%7D%20N%2FA%5E2)
B is the magnetic field which can be mathematically represented from the equation as
![B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }](https://tex.z-dn.net/?f=B%20%20%3D%20%20%5Csqrt%7B%20%5Cfrac%7B%202%20%2A%20%20%5Cmu_o%20%20%2A%20%20I%20%20%7D%7B%20c%7D%20%7D)
substituting values
![B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }](https://tex.z-dn.net/?f=B%20%20%3D%20%20%5Csqrt%7B%20%5Cfrac%7B%202%20%2A%20%204%5Cpi%20%2A10%5E%7B-7%7D%20%2A%20%20%200.0215%20%20%7D%7B%203.0%2A10%5E%7B8%7D%7D%20%7D)
![B = 1.342 *10^{-8} \ T](https://tex.z-dn.net/?f=B%20%20%3D%20%201.342%20%2A10%5E%7B-8%7D%20%5C%20%20T)
The area is mathematically represented as
![A = \pi r^2](https://tex.z-dn.net/?f=A%20%3D%20%20%5Cpi%20r%5E2)
substituting values
![A = 3.142 * (0.095)^2](https://tex.z-dn.net/?f=A%20%3D%20%203.142%20%2A%20%20%20%280.095%29%5E2)
![A = 0.0284](https://tex.z-dn.net/?f=A%20%3D%200.0284)
The angular velocity is mathematically represented as
![w = 2 * \pi * \frac{c}{\lambda }](https://tex.z-dn.net/?f=w%20%3D%20%202%20%2A%20%20%5Cpi%20%20%2A%20%20%5Cfrac%7Bc%7D%7B%5Clambda%20%7D)
substituting values
Generally the induced emf is mathematically represented as
![\epsilon = N * B * A * w * sin (wt )](https://tex.z-dn.net/?f=%5Cepsilon%20%20%3D%20%20N%20%2A%20%20B%20%20%2A%20%20A%20%20%2A%20%20w%20%2A%20sin%20%28wt%20%29)
At maximum induced emf ![sin (wt) = 1](https://tex.z-dn.net/?f=sin%20%28wt%29%20%20%3D%20%201)
So
![\epsilon = N * B * A * w](https://tex.z-dn.net/?f=%5Cepsilon%20%20%3D%20%20N%20%2A%20%20B%20%20%2A%20%20A%20%20%2A%20%20w)
substituting values
<em>12,25 km/h</em>
<em>≈ 3,4 m/s </em>
<em>v = d/t</em>
<em>= 12250m/h</em>
<em>= 12,25km/h</em>
<em>or</em>
<em>v = d/t</em>
<em>= 12250m/h</em>
<em>1h = 60m×60s = 3600s</em>
<em>= 12250m/3600s</em>
<em>≈ 3,4 m/s </em>
The correct answer is matter. Matter is a physical substance, so it is that only option provided that can be broken down to the subatomic particle level.
Answer:
The acceleration of the car will be ![a=9600m/sec^](https://tex.z-dn.net/?f=a%3D9600m%2Fsec%5E)
Explanation:
We have given that distance from stop sign s = 200 m
Time t = 0.2 sec
We have to find the constant acceleration
Now from second equation of motion ![s=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
![200=40\times 0.2+\frac{1}{2}\times a\times 0.2^2](https://tex.z-dn.net/?f=200%3D40%5Ctimes%200.2%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20a%5Ctimes%200.2%5E2)
![a=9600m/sec^](https://tex.z-dn.net/?f=a%3D9600m%2Fsec%5E)
So the acceleration of the car will be ![a=9600m/sec^](https://tex.z-dn.net/?f=a%3D9600m%2Fsec%5E)
3 is the answer to your question