Due to the fact that no one can consume .04 of a tablet, we can round down this answer to 1. This means that Mr. Jones should take C- 1 tablet per day.
I hope I've helped! :)
Answer:
A
Explanation:
A consensus is when you come to an agreement
Answer:
Current, I = 1000 A
Explanation:
It is given that,
Length of the copper wire, l = 7300 m
Resistance of copper line, R = 10 ohms
Magnetic field, B = 0.1 T
![\mu_o=4\pi \times 10^{-7}\ T-m/A](https://tex.z-dn.net/?f=%5Cmu_o%3D4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5C%20T-m%2FA)
Resistivity, ![\rho=1.72\times 10^{-8}\ \Omega-m](https://tex.z-dn.net/?f=%5Crho%3D1.72%5Ctimes%2010%5E%7B-8%7D%5C%20%5COmega-m)
We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :
![R=\rho \dfrac{l}{A}](https://tex.z-dn.net/?f=R%3D%5Crho%20%5Cdfrac%7Bl%7D%7BA%7D)
![R=\rho \dfrac{l}{\pi r^2}](https://tex.z-dn.net/?f=R%3D%5Crho%20%5Cdfrac%7Bl%7D%7B%5Cpi%20r%5E2%7D)
![r=\sqrt{\dfrac{\rho l}{R\pi}}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%5Cdfrac%7B%5Crho%20l%7D%7BR%5Cpi%7D%7D)
![r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%5Cdfrac%7B1.72%5Ctimes%2010%5E%7B-8%7D%5Ctimes%207300%7D%7B10%5Cpi%7D%7D)
r = 0.00199 m
or
![r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m](https://tex.z-dn.net/?f=r%3D1.99%5Ctimes%2010%5E%7B-3%7D%5C%20m%3D2%5Ctimes%2010%5E%7B-3%7D%5C%20m)
The magnetic field on a current carrying wire is given by :
![B=\dfrac{\mu_o I}{2\pi r}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B%5Cmu_o%20I%7D%7B2%5Cpi%20r%7D)
![I=\dfrac{2\pi rB}{\mu_o}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B2%5Cpi%20rB%7D%7B%5Cmu_o%7D)
![I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B2%5Cpi%20%5Ctimes%200.1%5Ctimes%202%5Ctimes%2010%5E%7B-3%7D%7D%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%7D)
I = 1000 A
So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.
"<span>The amount of matter in a substance" is the one among the following choices given in the question that best defines mass. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that has actually come to your desired help.</span>
A) 750 m
First of all, let's find the wavelength of the microwave. We have
is the frequency
is the speed of light
So the wavelength of the beam is
![\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bc%7D%7Bf%7D%3D%5Cfrac%7B3%5Ccdot%2010%5E8%20m%2Fs%7D%7B12%5Ccdot%2010%5E9%20Hz%7D%3D0.025%20m)
Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:
![y=\frac{m\lambda D}{a}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bm%5Clambda%20D%7D%7Ba%7D)
where
m = 1 since we are interested only in the central fringe
D = 30 km = 30,000 m
a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)
Substituting, we find
![y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B%281%29%280.025%20m%29%2830000%20m%29%7D%7B2.0%20m%7D%3D375%20m)
and so, the diameter is
![d=2y = 750 m](https://tex.z-dn.net/?f=d%3D2y%20%3D%20750%20m)
B) 0.23 W/m^2
First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is
![r=\frac{750 m}{2}=375 m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B750%20m%7D%7B2%7D%3D375%20m)
So the area is
![A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%20%3D%20%5Cpi%20%28375%20m%29%5E2%3D4.42%5Ccdot%2010%5E5%20m%5E2)
And since the power is
![P=100 kW = 1\cdot 10^5 W](https://tex.z-dn.net/?f=P%3D100%20kW%20%3D%201%5Ccdot%2010%5E5%20W)
The average intensity is
![I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BP%7D%7BA%7D%3D%5Cfrac%7B1%5Ccdot%2010%5E5%20W%7D%7B4.42%5Ccdot%2010%5E5%20m%5E2%7D%3D0.23%20W%2Fm%5E2)