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Marysya12 [62]
3 years ago
14

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t

he atmosphere and cause them to glow. What is the frequency of the circular orbit for an electron with speed 1.0×106m/s? Assume that the electron moves in a plane perpendicular to the magnetic field.
Physics
1 answer:
ollegr [7]3 years ago
5 0

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

r=\frac{mv}{qB}         (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

\omega=\frac{v}{r}

Then, you solve the equation (1) in order to obtain v/r:

\frac{v}{r}=\omega=\frac{qB}{m}

Finally, you replace the values of the parameters:

\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

f=2\pi \omega=5.5*10^7Hz

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A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.
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Answer:

39.7 m

Explanation:

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u is the velocity of the body at the beginning of that second

Solving for u, we find:

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Now we can call this velocity that we found v,

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h = s + s' = 23 m + 16.7 m = 39.7 m

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