When the rate of motion is the same direction the speed will decrease. True False

El conductor de un coche de 650 kg que va a 90 km/h frena y reduce su

Ivanshal [37]

A. 90km/h = 25 m/s
W=1/2mv^2
=1/2* 650* 25^2
= 203125(J)
b. v’= 90-50= 40km/h
= 100/9 m/s
W=1/2mv^2
=1/2*650*(100/9)^2
= 40123,45679

7 0

4 years ago

There are billions of stars in the universe how are they classified

Ilia_Sergeevich [38]

Answer: In 1926, an astronomer named Edwin Hubble decided to classify the galaxies, grouping them according to some logical scheme.

Explanation:

8 0

4 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.

x = -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.

x = +x0:

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

The honda insight, a hybrid electric vehicle, has an u.s. environmental protection agency (epa) gas mileage rating of 57 mi/gal

sergiy2304 [10]

Convert mL to gallons as follows:

355 mL = 0.355 L
0.355 L * (1 gal/3.785 L) = 0.0938 gallons

The equivalent km covered would be

57miles/gal * 0.0938 gal * (1.61 km/ 1 mile) = 8.61 kilometers

I hope I was able to help you. Have a good day!

3 0

4 years ago

A pendulum that was originally erected by Foucault at the Pantheon in Paris for the Paris Exhibition in 1851 was restored in 199

Morgarella [4.7K]

Answer:

13.1 m/s

Explanation: Given that a pendulum that was originally erected by Foucault at the Pantheon in Paris for the Paris Exhibition in 1851 was restored in 1995. It has a 28.0 kg sphere suspended from a 67.0-m light cable. If the amplitude of the swing is 5.00 m, what is the maximum speed of the sphere

The formula to use is :

W = sqrt ( L/g )

Where

L = 67m

g = 9.8m/s^2

A = 5m

Substitute all the parameters into the formula

W = sqrt ( 67 / 9.8 )

W = 2.6

To get the maximum velocity, we use the below formula.

V = wA

V = 2.6 × 5

V = 13.07 m/s

Therefore the maximum speed of the sphere is 13.1 m/s approximately.

7 0

3 years ago