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3 years ago

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A balloon with a charge of 6.0 μC is held a distance of 0.80 m from a second balloon having the same charge. Calculate the magni

tude of the repulsive force.

1 answer:

Yuliya22 [10]3 years ago

6 0

Answer:0.506 N

Explanation:

Given

Charge on first balloon $q_1=6\ \mu C$

Charge on second balloon is $q_2=6\ \mu C$

Distance between them $r=0.8\ m$

Electrostatic Repulsive force is given by

$F=\dfrac{kq_1q_2}{r^2}$

Where K is constant

$F=\dfrac{9\times 10^9\times 6\times 10^{-6}\times 6\times 10^{-6}}{(0.8)^2}$

$F=\dfrac{324\times 10^{-3}}{0.64}$

$F=0.506\ N$

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Can someone please help me out with this quiz will give brainiest and thanks to people

Virty [35]

Answer:

Energy transferred = 28.8 Joules.

1. Energy transferred = 144 Joules.

2. The unit of potential difference, volts can also be described as Joules per Coulombs.

3. Current, I = 6.945 Amperes.

Explanation:

<u>Part A.</u>

Given the following data;

Current, I = 1.2A

Time, t = 2 minutes

Potential difference, V = 12 volts.

To find the energy transfered;

Energy transferred = charge moved * potential difference

E = Q * V

Substituting into the equation, we have;

Energy transferred = (1.2 * 2) * 12

Energy transferred = 2.4 * 12

Energy transferred = 28.8 Joules.

<u>Part B.</u>

1. <em><u>Given the following data;</u></em>

Charge, Q = 24C

Potential difference = 6V

To find the energy transferred;

E = Q * V

Substituting into the equation, we have;

E = 24 * 6

E = 144 Joules.

2. Since we know that, Energy transferred = charge moved * potential difference

$Potential \; difference = \frac {Energy \; transferred}{Charged \; moved}$

The units of energy is Joules while the unit of the quantity of charge moved is Coulomb.

Therefore, the unit of potential difference becomes Joules per Coulomb.

3. <em><u>Given the following data;</u></em>

Potential difference = 18V

Energy transferred = 500J

Time, t = 4 minutes.

To find the current;

E = Q * V

Substituting into the equation, we have;

500 = Q*18

Q = 500/18

Q = 27.78C

But, Charge moved (Q) = current (I) * time (t)

Current, I = Q/t

Substituting into the equation, we have;

Current, I = 27.78/4

Current, I = 6.945 Amperes..

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2 years ago

A particle moves along the x axis so that its velocity at time t is given by v(t)=

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Answer:

a = 0.7267 , acceleration is positive therefore the speed is increasing

Explanation:

The definition of acceleration is

a = dv / dt

they give us the function of speed

v = - (t-1) sin (t² / 2)

a = - sin (t²/2) - (t-1) cos (t²/2) 2t / 2

a = - sin (t²/2) - t (t-1) cos (t²/2)

the acceleration for t = 4 s

a = - sin (4²/2) - 4 (4-1) cos (4²/2)

a = -sin 8 - 12 cos 8

remember that the angles are in radians

a = 0.7267

the problem does not indicate the units, but to be correct they must be m/s²

We see that the acceleration is positive therefore the speed is increasing

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2 years ago

Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a

victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

$F=\frac{GMm}{R^{2} }$

Where:

F => Gravitational force or force of attraction between two masses
M => Mass of asteroid 1
m => Mass of asteroid 2
R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

M=m

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

m=ρV

This is going to be helpful because the volume of a sphere is:

$\frac{4}{3}\pi r^{3}$

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

$F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }$
$F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }$
$F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2} }{36}$

Now let´s plug in the values we know:

$F = 1 lb$ mutual gravitational attraction force
$G = 6.67(10)^{-11}$ gravitational constant
$ρ_{iron} =491.5 \frac{lb}{ft^{3} }$

$1= \frac{6.67(10)^{-11} \pi ^{2} r^{4} (491.5)^{2}}{36}$

Solve for r and multiply by 2 because 2r = diameter

$d=2\sqrt[4]{\frac{1}{7.07(10)^{-5} } }$

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2 years ago

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a sp

Alex787 [66]

Answer:

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.

Explanation:

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2 years ago

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The force of attraction between the opposite charges of the ions in an ionic compound

Contact [7]

The force of attraction between the opposite charges of the ions in an ionic compound is an ionic bond.

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