A proton and an electron are placed in an electric field. Which undergoes the greater acceleration?

Physics

1 answer:

iren2701 [21]4 years ago

6 0

Newton's 2nd law:

Fnet = ma

Fnet is the net force acting on an object, m is the object's mass, and a is the acceleration.

The electric force on a charged object is given by

Fe = Eq

Fe is the electric force, E is the electric field at the point where the object is, and q is the object's charge.

We can assume, if the only force acting on the proton and electron is the electric force due to the electric field, that for both particles, Fnet = Fe

Fe = Eq

Eq = ma

a = Eq/m

We will also assume that the electric field acting on the proton and electron are the same. The proton and electron also have the same magnitude of charge (1.6×10⁻¹⁹C). What makes the difference in their acceleration is their masses. A quick Google search will provide the following values:

mass of proton = 1.67×10⁻²⁷kg

mass of electron = 9.11×10⁻³¹kg

The acceleration of an object is inversely proportional to its mass, so the electron will experience a greater acceleration than the proton.

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You are in a car traveling at 20 m/s. An ambulance is behind you traveling 35 m/s in the same direction. What frequency do you h

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Answer:

The frequency heard is 576.78 Hz

Explanation:

The Doppler effect is defined as the apparent frequency change of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave motion when the sender and receiver, or observer, move relative to each other.

This is what happens in the first part of this problem, where the sender is train A and the receiver is train B. They are both moving in opposite directions. In this case, where both are in motion, the frequency perceived by the receiver will increase when receiver and transmitter increase their separation distance and will decrease whenever the separation distance between them is reduced. The following expression is considered the general case of the Doppler effect:

$f'=f*\frac{v+-vR}{v+-vE}$

Where:

f ', f: Frequency perceived by the receiver and frequency emitted by the issuer respectively. Its unit of measurement in the International System (S.I.) is the hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s-1)

v: Velocity of propagation of the wave in the medium. It is constant and depends on the characteristics of the medium. In this case, the speed of sound in air is considered to be 343 m / s

vR, vE: Speed of the receiver and the emitter respectively. Its unit of measure in the S.I. is the m / s

±, ∓:

We will use the + sign:

In the numerator if the receiver approaches the emitter
In the denominator if the emitter moves away from the receiver

We will use the sign -:

In the numerator if the receiver moves away from the emitter
In the denominator if the emitter approaches the receiver

In this case you are in a car traveling at 20 m/s and an ambulance is behind you traveling 35 m/s in the same direction.

In this case the receiver, you in the car, moves away from the emitter, while the emitter, the ambulance, approaches the receiver behind you in the same direction. So the frequency is calculated by the expression:

$f'=f*\frac{v-vR}{v-vE}$