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4 years ago

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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What scientist was the first to propose the heliocentric model of the universe

Schach [20]

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3 years ago

The density of an object is dependent upon the object’s mass and ---

kotegsom [21]

Answer:Volume

Explanation:

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3 years ago

Choose the scenario in which the sound frequency of the waves is higher.

mrs_skeptik [129]

Answer:

B) the sound source moves towards you at 100 m/sec

Explanation:

The Dopper Effect is a phenomenon that occur when there is relative motion between an observer and a source of a wave. When this situation occurs, there is an apparent shift in frequency of the wave, as observed by the observer.

The apparent frequency observed by the observer is given by

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where

f is the original frequency of the wave

f' is the apparent frequency

v is the speed of the wave

$v_o$ is the velocity of the observer (positive if moving towards the source of the wave, negative otherwise)

$v_s$ is the velocity of the source (negative if moving towards from the observer, positive otherwise)

In this problem, we want to find the scenario in which the sound frequency is higher.

We see that in all 4 scenarios, the sound source is moving: this means we have to find the scenario in which the denominator of the equation is smaller.

First of all, we notice the sound source moves towards the observer, $v_s$ is negative, so the denominator is higher: this means that the correct option must be either A or B.

Also, we notice that since $v_s$ is negative, a value larger in magnitude will mean a smaller denominator: therefore, the correct answer will be

B) the sound source moves towards you at 100 m/sec

Since this situation will make the denominator of the formula the smallest possible.

5 0

3 years ago

An FM radio station broadcasts at 9.23 × 107 Hz. Given that the radio waves travel at 3.00 × 108 m/s, what is the wavelength of

STALIN [3.7K]

Answer:

3.2m

Explanation:

Given parameters:

Frequency of the FM radio = 9.23 x 10⁷Hz