Answer:
Kepler's third law of planetary motion
Explanation:
Kepler's third law of motion gives the relationship between a planet’s distance from the sun and the amount of time it takes the planet to orbit the sun. Its formula is given by :
![T^2=\dfrac{4\pi ^2}{GM}a^3](https://tex.z-dn.net/?f=T%5E2%3D%5Cdfrac%7B4%5Cpi%20%5E2%7D%7BGM%7Da%5E3)
Where
T is the time taken by the planet to orbit the sun
M is the mass of sun
G is the universal gravitational constant
a is the distance of planet form the sun
Answer:
Extraneous
Explanation:
Extraneous variables are any variables that you are not intentionally studying in your experiment or test
To answer the problem we would be using this formula which is I(peak) = P(peak)/(4πd^2) = 4.24413181578388 w/m^2
E = sqrt(I(peak)*Z0) = 39.9861614728793 V/m
B = µ0*sqrt(I(peak)/Z0) = 1.33379477721328E-7 T
(Free-space impedance Z0 = sqrt(µ0/e0) = 376.730313462204 ohms)
Answer
The current is 0.83 amps .
Option (C) is correct .
Explanation :
Formula for parallel resistors
![\frac{1}{R_{Parallel}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7BParallel%7D%7D%20%3D%20%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20%2B%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D)
As given
A combined circuit has two resistors in parallel (10.0 ohms and 14.0 ohms) .
![R_{1}= 10\ ohms](https://tex.z-dn.net/?f=R_%7B1%7D%3D%2010%5C%20ohms)
![R_{2}= 14\ ohms](https://tex.z-dn.net/?f=R_%7B2%7D%3D%2014%5C%20ohms)
Putting all the values in the above
![\frac{1}{R_{Parallel}} = \frac{1}{10} + \frac{1}{14}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7BParallel%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B10%7D%20%2B%20%5Cfrac%7B1%7D%7B14%7D)
L.C.M of (10,14) = 70
![\frac{1}{R_{Parallel}} = \frac{7+5}{70}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7BParallel%7D%7D%20%3D%20%5Cfrac%7B7%2B5%7D%7B70%7D)
![\frac{1}{R_{Parallel}} = \frac{12}{70}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7BParallel%7D%7D%20%3D%20%5Cfrac%7B12%7D%7B70%7D)
![R_{Parallel} = \frac{70}{12}](https://tex.z-dn.net/?f=R_%7BParallel%7D%20%3D%20%5Cfrac%7B70%7D%7B12%7D)
![R_{Parallel} = 5.83\ ohms\ (Approx)](https://tex.z-dn.net/?f=R_%7BParallel%7D%20%3D%205.83%5C%20ohms%5C%20%28Approx%29)
As given
Another resistors in the series is 5.0 ohms .
Formula
![R_{Total}=R_{Parallel}+ R_{3}](https://tex.z-dn.net/?f=R_%7BTotal%7D%3DR_%7BParallel%7D%2B%20R_%7B3%7D)
![R_{Parallel}=5.83\ ohms](https://tex.z-dn.net/?f=R_%7BParallel%7D%3D5.83%5C%20ohms)
![R_{3}=5\ ohms](https://tex.z-dn.net/?f=R_%7B3%7D%3D5%5C%20ohms)
![R_{Total}=5.83+5](https://tex.z-dn.net/?f=R_%7BTotal%7D%3D5.83%2B5)
![R_{Total}=10.83\ ohms](https://tex.z-dn.net/?f=R_%7BTotal%7D%3D10.83%5C%20ohms)
Now by using the ohms law .
![I = \frac{V}{R}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BV%7D%7BR%7D)
Where V is voltage , R is resistance and I is current .
As given
The power source is 9.0 volts .
V = 9 volts
![R_{Total}=10.83\ ohms](https://tex.z-dn.net/?f=R_%7BTotal%7D%3D10.83%5C%20ohms)
Putting values in ohms law
![I = \frac{9}{10.83}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B9%7D%7B10.83%7D)
I = 0.83 amps (Approx)
Therefore the current is 0.83 amps .
Option (C) is correct .