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elena-14-01-66 [18.8K]
4 years ago
8

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in t

he first 3.00μs3.00μs after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.
Physics
1 answer:
Softa [21]4 years ago
5 0

Answer:

Explanation:

A. Using

E= ma/q

E=m/q(2s/t²)

So

E= 9.1x10^-31/1.6*10^-19( 2*4.5/ 3*10-12)

E=5.7NC

The electric field has to be downward since the force is positive that is upward

B.

The electron acceleration is of the order of 10^11 times greater so for practical purposes we neglect the effect of gravity

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The motion of a mass on the end of a spring is a simple harmonic motion. In a simple harmonic motion, the total mechanical energy of the system is constant, and it is sum of the elastic potential energy (U) and the kinetic energy of the mass (K):

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As we see from the formula, since the total energy E is constant, when the displacement (x) increases, the speed (v) increases, and viceversa. Therefore, when the mass is at its equilibrium position (which corresponds to x=0), the velocity of the mass will be maximum.

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