Ask question

Ask question

All categories

3 years ago

14

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

al force between them

1 answer:

RUDIKE [14]3 years ago

4 0

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

$F = \frac{k|q_1||q_2|}{r^2}$

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

You might be interested in

Distinguish between nonsampling error and sampling error. Choose the correct answer below.

kiruha [24]

Answer:

The correct answer is D.

Non-sampling error is the error that results from under-coverage, non-response bias, response bias, or data-entry errors. Sampling error is the error that results because a sample is being used to estimate information about a population.

Explanation:

Sampling error is related to the variation between the true values of the sample and the population. If occurred, it is always random depending upon the sample chosen.

Non-sampling error can be random as well as non-random. Non-sampling error can occur irrespective of the sample chosen. It is related to the inappropriate analysis of the data.

5 0

3 years ago

Ciara is swinging a 0.015 kg ball tied to a string around her head in a flat, horizontal circle. The radius of the circle is 0.5

Marat540 [252]

Answer:

Option C

Explanation:

From the question we are told that:

Mass $m=0.015kg$

Radius $r=0.5m$

Time $t=0.70$

Generally the equation for Tension is mathematically given by

$T=\frac{4mr}{t^2} \pi$

$T=\frac{4*0.015*0.5}{(0.70)^2} (3.142)$

$T=0.60N$

Therefore

$T=0.60 N$ , toward the center of the circle

Option C

3 0

3 years ago

Help with this one please! Please give correct answers!

Verdich [7]

the answer is C because the air balloon is going down meaning that negative

5 0

4 years ago

A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational

a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: $F = 49.06N$

Pebble: $F = 29.44N$

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: $a =9.8m/s^{2}$

Pebble: $a =9.8m/s^{2}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{m1m2}{r^{2}}$ (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em> $m = 5.0 Kg$ <em>:</em>

m1 will be equal to the mass of the Earth $m1 = 5.972×10^{24} Kg$ and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth $r = 6371000m$ .

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}$

$F = 49.06N$

Newton's second law can be used to know the acceleration.

$F = ma$

$a =\frac{F}{m}$ (2)

$a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}$

$a =9.8m/s^{2}$

<em>Case for the pebble </em> $m = 3.0 Kg$ <em>:</em>

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}$

$F = 29.44N$

$a =\frac{F}{m}$

$a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}$

$a =9.8m/s^{2}$

3 0

3 years ago

Read 2 more answers

If the speed of an object does NOT change, the object is traveling at a

Elanso [62]

Answer:

If the speed does not change at all, the object would be moving at a <u>constant speed</u>.

5 0

3 years ago