B) Energy from the sun is the answer, I am sure;
The water cycle is driven primarily by the energy from the sun<span>
. This </span>
solar energy<span>
drives the cycle by evaporating water from the oceans, lakes, rivers, and even the soil. Other water moves from plants to the atmosphere through the process of transpiration.</span>
I = V / R
Current = (voltage) / (resistance)
Current = (500 V) / (250 ohms)
Current = (500/250) Amperes
<em>Current = 2 Amperes</em>
Answer:
The value is ![y = 3.097 * 10^{-5} \ m](https://tex.z-dn.net/?f=y%20%20%3D%203.097%20%2A%2010%5E%7B-5%7D%20%5C%20%20m%20)
Explanation:
From the question we are told that
The diameter of the pupil is ![d_p = 4.2 \ mm = 4.2 *10^{-3} \ m](https://tex.z-dn.net/?f=d_p%20%20%3D%20%204.2%20%5C%20mm%20%20%3D%20%204.2%20%2A10%5E%7B-3%7D%20%5C%20%20m)
The distance of the page from the eye ![d = 29 \ cm = 0.29 \ m](https://tex.z-dn.net/?f=d%20%3D%20%2029%20%5C%20%20cm%20%20%3D%20%200.29%20%5C%20%20m)
The wavelength is ![\lambda = 500 \ nm = 500 *10^{-9} \ m](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20%20500%20%5C%20nm%20%3D%20%20500%20%2A10%5E%7B-9%7D%20%5C%20%20m)
The refractive index is ![n_r = 1.36](https://tex.z-dn.net/?f=n_r%20%3D%20%201.36)
Generally the minimum separation of adjacent dots that can be resolved is mathematically represented as
![y = [ \frac{1.22 * \lambda }{d_p * n_r } ]* d](https://tex.z-dn.net/?f=y%20%20%3D%20%5B%20%5Cfrac%7B1.22%20%2A%20%20%5Clambda%20%7D%7Bd_p%20%2A%20n_r%20%7D%20%5D%2A%20d)
![y = [ \frac{1.22 * 500 *10^{-9} }{4.2 *10^{-3} * 1.36} ]* 0.29](https://tex.z-dn.net/?f=y%20%20%3D%20%5B%20%5Cfrac%7B1.22%20%2A%20%20500%20%2A10%5E%7B-9%7D%20%7D%7B4.2%20%2A10%5E%7B-3%7D%20%2A%201.36%7D%20%5D%2A%200.29)
![y = 3.097 * 10^{-5} \ m](https://tex.z-dn.net/?f=y%20%20%3D%203.097%20%2A%2010%5E%7B-5%7D%20%5C%20%20m%20)
By using Ohm's law, we can find what should be the resistance of the wire, R:
![R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega](https://tex.z-dn.net/?f=R%3D%20%5Cfrac%7BV%7D%7BI%7D%3D%20%5Cfrac%7B1.5%20V%7D%7B0.40%20A%7D%20%3D3.75%20%5COmega%20)
Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
![r=35 mm=0.35 \cdot 10^{-3} m](https://tex.z-dn.net/?f=r%3D35%20mm%3D0.35%20%5Ccdot%2010%5E%7B-3%7D%20m)
So the area is
![A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%20%3D%20%5Cpi%20%280.35%20%5Ccdot%2010%5E%7B-3%7D%20m%29%5E2%3D3.85%20%5Ccdot%2010%5E%7B-7%7D%20m%5E2)
And by using the resistivity of the Aluminum,
![\rho=2.65 \cdot 10^{-8} \Omega m](https://tex.z-dn.net/?f=%5Crho%3D2.65%20%5Ccdot%2010%5E%7B-8%7D%20%5COmega%20m)
, we can use the relationship between resistance R and resistivity:
![R= \frac{\rho L}{A}](https://tex.z-dn.net/?f=R%3D%20%5Cfrac%7B%5Crho%20L%7D%7BA%7D%20)
to find L, the length of the wire:
Current.A moving charge is an object that changes position to one particular obsever.