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Lubov Fominskaja [6]
3 years ago
12

An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 1.95 mT. If the speed of the

electron is 1.44 107 m/s, determine the following. (a) the radius of the circular path _____ cm (b) the time interval required to complete one revolution ____ s
Physics
1 answer:
zvonat [6]3 years ago
4 0

Answer: a) 42 *10^-3 m= 42 mm; b) 10.40 *10^-6 s=10.40 μs

Explanation: In order to response this problem we have to consider the Newton law for the circular movement,

Fm=m*ac where Fm is the magnetci force and ac the centripetal acceleration which is equal to v^2*R ( the raduis of teh curcular trajectory)

Fm=e*v*B considering that v and B are perpendicular

then we have:

e*v*B=m*v^2/R so

R=m*v/(e*B)= 9.1*10^-31*1.44*10^7/(1.6*10^-19*1.95*10^-3)= 42 mm

Then to calculate the time to complete one revolution ( period)

we know that ω=v*R and T= 2π/ω

then we have:

ω=1.44*10^7*42*10^-3=604.8 *10^3 rad/s

T=2*π/604.8 *10^3 = 10.4 μs

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3 years ago
Convert 5.5 kilometers into millimeters.​
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5500000 millimeters

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A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if
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To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

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F_p = 9*10^{22}N

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r = \frac{R}{2}

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F = \frac{GMm}{r^2}

Applying the new distance,

F = \frac{GMm}{(\frac{R}{2})^2}

F =  4\frac{GMm}{R^2}

Replacing with the previous force,

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Replacing our values

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