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3 years ago

Gravity is a force of attraction that acts

Physics

1 answer:

bixtya [17]3 years ago

5 0

Answer:gravity is a force of attraction that acts between two masses separated by distance

Explanation:

Gravity is a force of attraction that acts between two masses separated by distance

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How does the earth orbit the sun?

scoray [572]

Answer:

The Sun's gravity pulls on the planets, just as Earth's gravity pulls down anything that is not held up by some other force and keeps you and me on the ground.

Explanation:

Hope that helps

6 0

3 years ago

A ball starts from rest and rolls down a hill at a constant acceleration of 5 m/s2. If it travels for 8 m how fast is it going i

makkiz [27]

Hi there!

We can use the following kinematic equation:

$v_f^2 = v_i^2 + 2ad$

vf = final velocity (? m/s)
vi = intial velocity (0 m/s)

a = acceleration (5 m/s²)

d = displacement (8 m)

Plug in the givens and solve.

$v_f^2 = 0 + 2(5)(8)\\\\v_f = \sqrt{80} = \boxed{8.944 \frac{m}{s}}$

5 0

3 years ago

Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o

VikaD [51]

Answer:

For the point inside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

For the point outside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}$

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}$

We have to integrate it over the ring, which is an angular integration.

$E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}$

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

We will only change the boundaries of the last integration.

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

6 0

3 years ago

5.0 Points The blue color of the sky is the result of

mina [271]

The blue color of the sky is the result of scattering. The answer is letter B. it is specifically called the Rayleigh scattering. This blue color is the result of the incoming rays of the Sun into the earth’s atmosphere. It has he lowest wavelength which as we all see is the blue spectrum of light.

7 0

3 years ago

Which is not true of the Intertropical Convergence Zone?A) It features heavy precipitation B) It's where the trade winds collide

olasank [31]

<h2>Answer: C) It's a high-pressure zone with sinking air</h2>

Explanation:

The intertropical convergence zone is the region of the terrestrial globe where the trade winds of the northern hemisphere converge with those of the southern hemisphere.

It is characterized by being <u>a belt of low pressure</u> and inconsistent location around the equator constituted by ascending air currents, where large masses of warm and humid air converge from the north and south of the intertropical zone.

The reason of its inconsistent location is due to the movements of the Earth with the seasons, having as a consequence the amount variation of heat energy from the sun in this region.

7 0

3 years ago