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goldfiish [28.3K]

2 years ago

5

Two charged particles are placed 2.0 meters apart. The first charge is +2.0 E-6 C, and the second charge is +4.0 E-6 C. What is

the electrical force between the two charges? (5 points)
(k = 9.0 E9 Nm2/c2)
+1.8 E-2 N and it is repulsive
+3.6 E-2 N and it is repulsive
+4.5 E-2 N and it is attractive
+9.0 E-2 N and it is attractive

1 answer:

Anestetic [448]2 years ago

8 0

The electrical force between the two charges can be calculated as +1.8×10⁻² N and its repulsive.

To find the electrical force, distance = 2 meters

charge q1 = 2×10⁻⁶ C

charge q2 = 4×10⁻⁶ C

<h3>Define coulomb's law and solve with formula.</h3>

The force of attraction or repulsion acting along a straight line between two electric charges is directly proportional to the product of the charges and inversely to the square of the distance between them.

Formula can be written as,

F = K ( q1q2 / r² )

F - electric force

k - Coulomb constant

q1, q2 - charges

r - distance of separation

Substituting the values in the formula,

F = 9 × 10⁹ Nm²/C² ( (2×10⁻⁶ C × 4×10⁻⁶ C) / (2²))

= 0.0179751

F = 1.8 × 10 ⁻² N.

As both the charges q1 and q2 are positive, the charges gets repulsive.

SO, the correct answer is Option A.

Learn more about coulomb's law,

brainly.com/question/506926

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Estimate the kinetic energy of the earth with respect to the sun as the sum of two terms.

nekit [7.7K]

The definition of kinetic energy allows to find the result for the relationship between the energy of the sun and the Earth is:

The kinetic energy ratio is $\frac{K_{Sum} }{K_{Earth}} = 5.3 \ 10^2$

<h3 /><h3 /><h3> Kinetic enrgy.</h3>

Kinetic energy is the energy due to the movement of bodies, it is given by the relation

K = ½ m v²

where K is the kinetic energy, m the mass of the body and v the velocity of the body.

In a compound motion it is common to separate energy into parts to simplify calculations.

Translational kinetic energy. Due to the linear movement of the body

$K_{tras} =\frac{1}{2} m v^2$

Rotational kinetic energy. Due to the rotational movement of the body.

$K_{rot} = \frac{1}{2} I w^2$

Where I is the inrtia momentum and w the angular velocity.

They indicate that we compare the kinetic energy of the sun and the Earth.

The Earth has two movements, one of rotation about its axis with a period of T = 24 h and one of translation with respect to the Sun with a period of T= 365 days, therefore the kinetic energy of the Earth.

$K_{earth} = K_{tras} + K_{rot}$

Linear and rotational speed are related.

v = w r

The Earth is an almost spherical body therefore the moment of inertia of a solid sphere.

I = $\frac{2}{5 } m r^2$

Let's subatitute.

$K_{earth} = \frac{1}{2} \ m r^2_{tras} w^2_{tras} + \frac{1}{2} ( \frac{2}{5} m r^2_{earth}) w^2_{rot}$

The movement of the Earth around the sun is almost circular, therefore we can use the relations of the uniform circular movement, where the angle for one revolution is 2π radians and the time is called the period.

$w = \frac{2 \pi}{T}$

Let's substitute.

$K_{earth} = \frac{1}{2} m ( \frac{2\pi r^2_{tras}}{T_{tras}})^2 \ + \frac{1}{5} m (\frac{2\pi r^2_{earth} }{T^2_{rot}})^2$

$K_{earth} = 4 \pi^2 \ m \ ( \frac{1}{2} [ \frac{r_{tras}}{T_{tras}y} ]^2 + \frac{1}{5} [ \frac{r_{rot}}{T_{rot}}]^2)$

Data for Earth are tabulated:

Mass m = 5.98 1024 kg

Radius r = 6.37 10⁶ m

Radius orbits tras = 1.496 10¹¹ m

Rotation period $T_{rot}$ = 24 h ( $\frac{3600s}{1h}$ ) = 8.64 10⁴s

Translation period $T_{tras}$ = 365 d ( $\frac{24h}{1 d}$ ) ( $\frac{3600s}{1h}$ ) = 3.15 10⁷ s

Let's calculate.

$K_{earth} = 4 \pi^2 5.98 \ 10^{24} ( \frac{1}{2} ( \frac{1.496 \ 10^{11}}{3.15 \ 10^7 } )^2 \ + \frac{1}{5}( \frac{6.37 \ 10^6 }{8.64 \ 10^4})^2 )$

$K_{earth} = 2.36 \ 10^{26 } \ (1.128 \ 10^7 + 1.087 \ 10^3)$

$K_{earth}= 2.66 \ 10^{33} J$

Let's analyze the kinetic energy for the Sun, this is inside the solar system therefore it has no translation movement and is approximately a sphere with a rotation period of $T_{Sum}$ = 27 days.

The kinetic energy of the sun is;

$K_{sum} = K_{rot} = \frac{1}{2} I w^2$

$K_{sum} = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{2\pi}{T_{sum}})^2$

$K_{sum} = \frac{4\pi^2 }{5} M (\frac{R}{T_{rot}})^2$

The tabulated data for the sun are:

Mass m = 1,991 1030 kg.

Radius R = 6.96 10⁸ m

Period T = 27 d ( $\frac{24h}{1 d}$ ) ( $\frac{3600s}{1h}$ ) = 2.33 10⁶ s

Let's calculate.

$K_{sum} = 1.40 \ 10^{36} J$

The relationship of the kinetic energy of the sun and the Earth is:

$\frac{K_{sum}}{K_{earth}} = \frac{1.40 \ 10^{36}}{2.66 \ 10^{33}}$

$\frac{K_{sum}}{K_{earth}} = 5.3 \ 10^2$

In conclusion using the definition of kinetic energy we can shorten the result for the relationship between the energy of the sun and the Earth is:

The kinetic energy ratio is: $\frac{K_{Sum}}{K_{Earth}} = 5 \ 10^2$

Learn more about kinetic energy here: brainly.com/question/25959744

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Convection is the circular motion that occurs as hotter air or liquid increases when the cooler air or liquid drops down, and has faster moving molecules, rendering it less dense. Convection currents within the earth shift layers of magma, and currents are formed by convection in the ocean.

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A wave with a wavelength of 15m travel at 330m/s.Calculate its frequency.

katovenus [111]

ANSWER:

22 Hz.

Explanation:

Frequency = (speed) divided by (wavelength) = (330) / (15) =

8 0

3 years ago

The demand for your hand-made skateboards, in weekly sales, is q = −3p + 700 if the selling price is $p. You are prepared to sup

Y_Kistochka [10]

Answer:

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Explanation:

The price p to sell your skateboards for so that there is neither a shortage nor a surplus is the price that makes equal the quantity of sales and the quantity of supply, so p is equal to:

q (sales) = q (supply)

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On the graph of voltage versus current, which line represents a 2.0 Ω resistor?

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Answer:

<h2>line B</h2>

Explanation:

According to ohm's law V = IR where;

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In order to calculate the line that is equal to 2ohms, we need to calculate the slope of each line using the formula.

For line B, R = ΔV/ΔI

R = V₂-V₁/I₂-I₁

R = 14.0-4.0/7.0-2.0

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