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Juli2301 [7.4K]
3 years ago
10

An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels fo

r a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
A. The time the electron is in the magnetic field
B. The magnitude of the net force acting on the electron inside the field
C. The magnitude of the electron's acceleration inside the field
D. The radius of the circular path the electron travels
Physics
1 answer:
pogonyaev3 years ago
4 0

Answer:

C. The magnitude of the electron's acceleration inside the field

D. The radius of the circular path the electron travels

Explanation:

The radius of the electron's motion in a uniform magnetic field is given by

R = \frac{MV}{qB}

where;

m is the mass of the electron

q is the charge of the electron

B is the magnitude of the magnetic field

V is speed of the electron

R is the radius of the electron's

Thus, the radius of the of the electron's motion will change since it depends on speed of the electron.

The magnitude of the electron's acceleration inside the field  is given by;

a_c = \frac{V^2}{R}

where;

a_c is centripetal acceleration of electron

Thus, the magnitude of the electron's acceleration inside the field will change since it depends on the electron speed.

The time the electron is in the magnetic field is given by;

T = \frac{2\pi M}{qB}

The time of electron motion will not change

The magnitude of the net force acting on the electron inside the field will not change;

qVB = \frac{MV^2}{R} \\\\qVB - \frac{MV^2}{R}  = 0

Therefore, the correct options are "C" and "D"

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Answer:

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Explanation:

Hello

the same units of measure should be used, then

Step 1

convert  42 m/s ⇒   km/h

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Step 2

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now, isolating s

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and replacing

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Have a great day

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Explanation:

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Explanation:

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Answer:

h_m=25.8\ m

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up without taking into consideration friction with the air.

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\mathbf{h_m=25.8\ m}

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