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2 years ago

11

A 1,500-kg car accelerates along a flat track to a speed of 20 m/s. how much does it's gravitational potential energy increases

over the length of the track?

1 answer:

lesya [120]2 years ago

3 0

Look at bitesizes physics section, they have all the information you need to complete this question.

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White visible light travels from the Sun to Earth. What is the best description of white light?

aliya0001 [1]

Answer:

As investigated by Sir Isaac Newton,the white light contains all colours of light.

This shorten by a term rainbow colours.

6 0

2 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.

<u>x = -x0:</u>

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.

<u>x = +x0: </u>

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

Which type of galaxy has the most active star formation?

UkoKoshka [18]

That answer is spiral galaxies

8 0

3 years ago

Read 2 more answers

What is polarization?

worty [1.4K]

Answer:

if your glasses are polarized, you can see the fish in the water. also im pretty sure its d

Explanation:

5 0

3 years ago

What would the answer be ?

Marianna [84]

The second one since you’re changing the soil up by adding different fertilisers. This will be you’re independent variable. And you’re dependent variable is your result = the plant height .
Hope this helps :)

4 0

3 years ago