Answer:
C
Step-by-step explanation:
The charge of the electrons is
.
The given parameters:
- <em>Number of revolution of the electron, N = 1.4 x 10⁶ rev per sec</em>
- <em>Flux density, B = 10⁻⁵ T</em>
- <em>Mass of electron, m = 9.11 x 10⁻³¹ kg</em>
<em />
The angular speed of the electron is calculated as follows;
![\omega = 2\pi N\\\\\omega = 2\pi \ \frac{rad}{1 \ rev} \times 1.4 \times 10^6 \ rev/s\\\\\omega = 8.798 \times 10^6 \ rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20N%5C%5C%5C%5C%5Comega%20%3D%202%5Cpi%20%5C%20%5Cfrac%7Brad%7D%7B1%20%5C%20rev%7D%20%5Ctimes%201.4%20%5Ctimes%2010%5E6%20%5C%20rev%2Fs%5C%5C%5C%5C%5Comega%20%3D%208.798%20%5Ctimes%2010%5E6%20%5C%20rad%2Fs)
The charge of the electrons is calculated as follows;
![F_c= m\omega ^2 r\\\\F_{B}= qvB = q\omega r B \\\\F_c = F_B\\\\m\omega ^2 r = q\omega r B \\\\q = \frac{m\omega ^2 r}{\omega r B} \\\\q = \frac{m \omega }{B} \\\\q = \frac{9.11 \times 10^{-31} \times 8.798 \times 10^6}{10^{-5}} \\\\q = 8.02 \times 10^{-19} \ C](https://tex.z-dn.net/?f=F_c%3D%20m%5Comega%20%5E2%20r%5C%5C%5C%5CF_%7BB%7D%3D%20qvB%20%3D%20q%5Comega%20r%20B%20%5C%5C%5C%5CF_c%20%3D%20F_B%5C%5C%5C%5Cm%5Comega%20%5E2%20r%20%3D%20%20q%5Comega%20r%20B%20%5C%5C%5C%5Cq%20%3D%20%5Cfrac%7Bm%5Comega%20%5E2%20r%7D%7B%5Comega%20r%20B%7D%20%5C%5C%5C%5Cq%20%3D%20%5Cfrac%7Bm%20%5Comega%20%7D%7BB%7D%20%5C%5C%5C%5Cq%20%3D%20%5Cfrac%7B9.11%20%5Ctimes%2010%5E%7B-31%7D%20%5Ctimes%208.798%20%5Ctimes%2010%5E6%7D%7B10%5E%7B-5%7D%7D%20%5C%5C%5C%5Cq%20%3D%208.02%20%5Ctimes%2010%5E%7B-19%7D%20%5C%20C)
The charge of the electrons is
.
The complete question here:
Electrons execute 1.4*10^6 revolutions per sec in a magnetic field of flux density 10⁻⁵ T. What will be the value of charge of the electron?
Learn more about magnetic force of electrons here: brainly.com/question/3990732
Answer:
A. No, this is not a valid inference because he asked only 20 families
Step-by-step explanation:
You cannot assume that those are the only families that would want to go; you have to ask all the families.
393216 times 4 because you could continue it on until u get to nine
The point of observation is 2500 m away from the foot of the building.
The angle of elevation is 4°.
We need to find the height 'h' of the building.
With respect to 4°,
2500 is the adjacent side.
'h' is the opposite side.
The trigonometric ratio associating opposite & adjacent is tan.
We have
![tan\theta =\frac{opposite}{adjacent}](https://tex.z-dn.net/?f=%20tan%5Ctheta%20%3D%5Cfrac%7Bopposite%7D%7Badjacent%7D%20)
![tan4^{o}=\frac{h}{2500}](https://tex.z-dn.net/?f=%20tan4%5E%7Bo%7D%3D%5Cfrac%7Bh%7D%7B2500%7D%20)
Cross multiplying we get
h = 2500 tan4°
h= 174.82 m
Option B) is the right answer.