Always true everthing has gravity
Answer:
77647
Explanation:
= Efficiency = 34%
Power used in 1 home = 0.02 MW
Total power is
![P=\dfrac{800}{\eta}\\\Rightarrow P=\dfrac{800}{0.34}\\\Rightarrow P=2352.94117\ MW](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7B800%7D%7B%5Ceta%7D%5C%5C%5CRightarrow%20P%3D%5Cdfrac%7B800%7D%7B0.34%7D%5C%5C%5CRightarrow%20P%3D2352.94117%5C%20MW)
Waste of power
![2352.94117-800=1552.94117\ W](https://tex.z-dn.net/?f=2352.94117-800%3D1552.94117%5C%20W)
Number of homes would be given by
![n=\dfrac{1552.94117}{0.02}=77647.0585\ homes](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7B1552.94117%7D%7B0.02%7D%3D77647.0585%5C%20homes)
The number of homes that could be heated with the waste heat of this one power plant is 77647
Complete Question
You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced
meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are Ï out of phase with each other (that is to say, Ao,- ). If you observe that on a screen placed 4 meters from the two slits that the distance between the bright spot closest to center of the pattern is 1.5 cm, what is the wavelength of the laser?
Answer:
The wavelength is ![\lambda = 56250 nm](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20%2056250%20nm)
Explanation:
From the question we are told that
The distance of slit separation is
The distance of the screen is ![D = 4 \ m](https://tex.z-dn.net/?f=D%20%3D%20%204%20%5C%20%20m)
The distance between the bright spot closest to the center of the interference is ![k = 1.5 \ cm = 0.015 \ m](https://tex.z-dn.net/?f=k%20%20%20%3D%201.5%20%5C%20cm%20%3D%200.015%20%5C%20%20m)
Generally the width of the central maximum fringe produced is mathematically represented as
![y = 2 * k = \frac{ D * \lambda}{d}](https://tex.z-dn.net/?f=y%20%20%3D%20%202%20%2A%20%20k%20%20%3D%20%5Cfrac%7B%20D%20%20%2A%20%20%5Clambda%7D%7Bd%7D)
=> ![2 * 0.015 = \frac{ \lambda * 4}{ 7.5 *10^{-3}}](https://tex.z-dn.net/?f=2%20%2A%20%200.015%20%3D%20%20%5Cfrac%7B%20%5Clambda%20%20%2A%20%204%7D%7B%207.5%20%2A10%5E%7B-3%7D%7D)
=> ![\lambda = 56250 *10^{-9} \ m](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20%2056250%20%2A10%5E%7B-9%7D%20%5C%20m)
=> ![\lambda = 56250 nm](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20%2056250%20nm)
The area-
The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-blue triangle and the dark-blue rectangle.
<span><span>Area of light-blue triangle -
<span>The width of the triangle is 4 seconds and the height is 8 meters per second. To find the area, you use the equation: <span>area of triangle = 1⁄2 × base × height </span><span>so the area of the light-blue triangle is 1⁄2 × 8 × 4 = 16m. </span></span></span><span> Area of dark-blue rectangle
The width of the rectangle is 6 seconds and the height is 8 meters per second. So the area is 8 × 6 = 48m.</span><span> Area under the whole graph
<span>The area of the light-blue triangle plus the area of the dark-blue rectangle is:16 + 48 = 64m.<span>This is the total area under the distance-time graph. This area represents the distance covered.</span></span></span></span>