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3 years ago

The magnitude of the electric force between two protons is 2.3x10-26 n. how far apart are they?

1 answer:

6 0

To solve this problem, we make use of the Coulumb’s law which relates the electrical force to the charges. The formula used in Coulumb’s law is given as:

F = k q1 q2 / r^2

where,

F = electric force = 2.3 * 10^-26 N

k = Coulumb’s constant = 9 * 10^9 N m^2/C^2

q1 = q2 = electric charge of protons = 1.602 * 10^-19 C

r = distance between the two protons

Substituting the given values:

2.3 * 10^-26 N = (9 * 10^9 N m^2/C^2) (1.602 * 10^-19 C)^2 / r^2

r^2 = 0.01 m^2

r = 0.1 m

Therefore the protons are only 0.1 m apart or 10 cm.

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Standing waves are set up on two strings fixed at each end, as shown in the drawing. The two strings have the same tension and m

uysha [10]

The beat frequency produced by the two standing waves is 13 Hz.

<h3>The wavelength of the shorter string</h3>

The wavelength of the shorter string is calculated as follows;

$L = \frac{\lambda}{2} \\\\\lambda = 2L\\\\\lambda = \frac{v}{f} \\\\\lambda = \frac{41.9}{225} \\\\\lambda = 0.186 \ m\\\\\lambda = 18.6 \ cm\\\\L= \frac{\lambda }{2} \\\\L = \frac{18.6 \ cm}{2} = 9.3\ cm$

<h3>The length of the longer string</h3>

$L_2 = 0.58 \ cm \ + 9.3 \ cm\\\\L_2 = 9.88 \ cm \\\\\lambda _2 = 2L_2\\\\\lambda _2 = 2(9.88 \ cm)\\\\\lambda_2 = 19.76 \ cm = 0.1976 \ m$

The frequency of the longer string is calculated as follows;

$v_1 = v_2\\\\f_2 = \frac{v_2}{\lambda_2} \\\\f_2 = \frac{41.9}{0.1976} \\\\f_2 = 212 \ Hz$

<h3>Beat frequency</h3>

The beat frequency produced by the two standing waves is calculated as follows;

$F_b = 225 \ Hz \ - \ 212 \ Hz\\\\F_b = 13 \ Hz$

Learn more about beat frequency here: brainly.com/question/3086912

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3 years ago

Many students use earbuds or headphones while working on a computer or listening to music. Electromagnetics are within the liste

Katyanochek1 [597]

Answer:

When the electrical current passes through the conductors, the polarity is changed, and the diaphragm interacts with the permanent magnets, vibrating and creating sound.

Explanation:

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3 years ago

What type of data do scientists use to predict tornadoes?

Sav [38]

Correct answer would be d

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3 years ago

What is atmospheric pressure ? How can we calculate it?

oksian1 [2.3K]

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3 years ago

A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a

Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

$Z_{R}$ = R

R = resistance

$Z_{L}$ = jωL

ω = voltage source angular frequency, L = inductance

$Z_{C}$ = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

$Z_{R}$ = 217Ω

$Z_{L}$ = j(220)(0.875) = j192.5Ω

$Z_{C}$ = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$ + $Z_{C}$

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

$v_{R}$ = I× $Z_{R}$

$v_{L}$ = I× $Z_{L}$

$v_{C}$ = I× $Z_{C}$

Given values:

I = (0.0569∠1.147+220t)A

$Z_{R}$ = 217Ω = (217∠0)Ω

$Z_{L}$ = j192.5Ω = (192.5∠π/2)Ω

$Z_{C}$ = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

$v_{R}$ = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

$v_{L}$ = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

$v_{C}$ = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

$v_{R}$ = (12.35∠1.147+220t)V

$v_{L}$ = (10.95∠2.718+220t)V

$v_{C}$ = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

$v_{R}$ = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

$v_{L}$ = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

$v_{C}$ = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0

3 years ago