Answer:
The answer are B
Explanation:the two samples haves yellow color :)
Answer:
D
Explanation:
At noon on the summer solstice. The sun dagger was used to observe astrological readings in the ancient time
E S *
The "E" represents Earth, "S" represent Sun, and the "*" represents the nearest star(which is Proxima Centauri).
The main thing to worry about here is units, so ill label everything out.
D'e,s'(Distance between earth and sun) = .<span>00001581 light years
D'e,*'(Distance between earth and Proxima) = </span><span>4.243 light years
Now this is where it gets fun, we need to put all the light years into centimeters.(theres alot)
In one light year, there are </span>9.461 * 10^17 centimeters.(the * in this case means multiplication) or 946,100,000,000,000,000 centimeters.
To convert we multiply the light years we found by the big number.
D'e,s'(Distance between earth and sun) = 1.496 * 10^13 centimeters<span>
D'e,*'(Distance between earth and Proxima) = </span><span>4.014 * 10^18 centimeters
</span>
Now we scale things down, we treat 1.496 * 10^13 centimeters as a SINGLE centimeter, because that's the distance between the earth and the sun. So all we have to do is divide (4.014 * 10^18 ) by (<span>1.496 * 10^13 ).
Why? because that how proportions work.
As a result, you get a mere 268335.7 centimeters.
To put that into perspective, that's only about 1.7 miles
A lot of my numbers came from google, so they are estimations and are not perfect, but its hard to be on really large scales.</span>
Answer:
a) 1.6*10^6 V
b) 13.35*10^6 V
Explanation:
The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:
(1)
q1 = 3.90µC = 3.90*10^-6 C
q2 = -2.4µC = -2.4*10^-6 C
r1 = 1.25 cm = 0.0125 m
r2 = -1.80 cm = -0.018 m
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
You replace all the parameters in the equation (1):
![V=k[\frac{q_1}{r_1}+\frac{q_2}{r_2}]\\\\V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0125m}+\frac{-2.4*10^{-6}C}{0.018m}]=1.6*10^6V](https://tex.z-dn.net/?f=V%3Dk%5B%5Cfrac%7Bq_1%7D%7Br_1%7D%2B%5Cfrac%7Bq_2%7D%7Br_2%7D%5D%5C%5C%5C%5CV%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5B%5Cfrac%7B3.90%2A10%5E%7B-6%7DC%7D%7B0.0125m%7D%2B%5Cfrac%7B-2.4%2A10%5E%7B-6%7DC%7D%7B0.018m%7D%5D%3D1.6%2A10%5E6V)
hence, the total electric potential is approximately 1.6*10^6 V
b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:
r1 = 0.0150m - 0.0125m = 0.0025m
r2= 0.015m + 0.018m = 0.033m
Then, you replace in the equation (1):
![V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0025m}+\frac{-2.4*10^{-6}C}{0.033m}]=13.35*10^6V](https://tex.z-dn.net/?f=V%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5B%5Cfrac%7B3.90%2A10%5E%7B-6%7DC%7D%7B0.0025m%7D%2B%5Cfrac%7B-2.4%2A10%5E%7B-6%7DC%7D%7B0.033m%7D%5D%3D13.35%2A10%5E6V)
hence, for y = 1.50cm you obtain V = 13.35*10^6 V
