Please help me!<br>each form of a characteristic is called what?

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds.

salantis [7]

Answer:

130m

Explanation:

You just have to multiply velocity by the time traveled:

100m/s * 1.3s = 130m!

5 0

3 years ago

Read 2 more answers

Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .

Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

Where;

$G$ is the Gravitational Constant and its value is $6.674(10^{-11})\frac{m^{3}}{kgs^{2}}$

$M=1.9(10^{27})kg$ is the mass of Jupiter

$a=4.22(10^{5})km=4.22(10^{8})m$ is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}$

$T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}$

$T=\sqrt{2.339(10^{10})s^{2}}$

Then:

Which is the same as:

Therefore, the answer is:

The orbital period of Io is 42.482 h

7 0

3 years ago

Which would most likely cause specular reflection? O a pathway with rough rocks a shiny, smooth leaf a small patch of soil a rou

andrew11 [14]

Answer:

a shiny smooth leaf

Explanation:

A shiny smooth leaf will cause specular reflection. Other choices will cause diffused reflection from the surface.

A specular reflection is similar to how a mirror or smooth surface reflects. The incident light is given off as a single ordered reflection from the surface of a body.

For this to occur, the surface incident must be smooth and without rough patterns on it.

A path way with rough rocks, small patch of soil and rough logs will give off diffused reflection

8 0

3 years ago

Read 2 more answers

The following are the Earth–Sun distance at the equinoxes and solstices: March equinox 149.0 million km June solstice 152.0 mill

Mice21 [21]

Answer:

During <u>winter (late December/early January)</u> the Earth is closest to the Sun and during <u>summer (late June/early July)</u> the Earth is farthest from the Sun.

Explanation:

In the northern hemisphere, the earth usually comes closer to the sun during the time of winter season, mostly in late December or early January.

On the other hand, the earth is farthest from the sun during the time of summer season, mostly in late June or early July.

When the earth is closer to the sun, during the winter, it is comparatively cold. It is due to the absorption of a lesser amount of incoming solar radiation. The tilt of the earth is also responsible for this low temperature.

But, when the earth is farthest from the sun, during the summer, it is comparatively hot. It is due to the absorption of a large amount of incoming solar radiation.

5 0

3 years ago

Please help me!each form of a characteristic is called what?