Answer:
It is made up of iron, steel or any other ferromagnetic substance or ferromagnetic composite, having permanent magnetic properties. The magnet has two poles: a north and a south pole. When you suspend it freely, the magnet aligns itself so that the north pole points towards the magnetic north pole of the earth.
Index of refraction can be calculated using the following formula:
![n=\frac{c}{v}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7Bc%7D%7Bv%7D)
where
in a vacuum and "n" is the index of refraction. So we must calculate our velocity by manipulating the index of refraction formula and isolate v and so:
![v=\frac{c}{n}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bc%7D%7Bn%7D)
We could go one by one and calculate the velocity or use logic. If you divide by a larger number the output will be a smaller number. Therefore, the answer would be n=1.62 is the slowest.
![v=\frac{3 \times 10^8 \frac{m}{s}}{1.62}= 1.85185 \times 10^8 \frac{m}{s}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B3%20%5Ctimes%2010%5E8%20%5Cfrac%7Bm%7D%7Bs%7D%7D%7B1.62%7D%3D%201.85185%20%5Ctimes%2010%5E8%20%5Cfrac%7Bm%7D%7Bs%7D)
By that logic when n=1 light would travel the fastest of these four scenarios. The key thing to understand about these problems is that the speed of light slows down as it goes through a medium, in a vacuum it's 3 x 10^ 8 m/s but as it goes through say air or water light slows down.
I think that the answer to this is a. hope this helps
Initial velocity u = 20 m/s
Initial horizontal velocity = 20 cos30° = 20 * 0.866 = 17.32 m/sec.
Initial vertical velocity = 20 sin30° = 20 * 1/2 = 10 m/sec.
time taken t = u/g = 10/10 =1 sec. ( approximating g to 10m/sec^2)
Maximum height h = ut + 1/2 * g * t^2
h = 10 *1 - 1/2 * 10 * 1* 1
h = 10 - 5 = 5 metres.
Total time in air = 2t = 2 seconds.
This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.
![I=\frac{Q}{t}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BQ%7D%7Bt%7D)
Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.
![\begin{gathered} I=\frac{2.0\times10^{-4}C}{2.0\times10^{-6}\sec } \\ I=1.0\times10^{-4-(-6)}A \\ I=1.0\times10^{-4+6}A \\ I=1.0\times10^2A \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20I%3D%5Cfrac%7B2.0%5Ctimes10%5E%7B-4%7DC%7D%7B2.0%5Ctimes10%5E%7B-6%7D%5Csec%20%7D%20%5C%5C%20I%3D1.0%5Ctimes10%5E%7B-4-%28-6%29%7DA%20%5C%5C%20I%3D1.0%5Ctimes10%5E%7B-4%2B6%7DA%20%5C%5C%20I%3D1.0%5Ctimes10%5E2A%20%5Cend%7Bgathered%7D)
<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>
<em />
<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>