1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Andrei [34K]
3 years ago
14

Large amounts of water that quickly infiltrate soil on a sloped surface can

Physics
1 answer:
sergey [27]3 years ago
7 0
A is the correct choice
You might be interested in
An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
A 100kg crate slides along a floor with a starting velocity of 21ms If the force due to friction is 8N how long will it take for
GaryK [48]

A 100kg crate slides along a floor with a starting velocity of 21 m/s. If the force due to friction is 8N, then, it will take 262.5 s for the box to come to rest.

We'll begin by calculating the declaration of the box. This can be obtained as follow:

Force (F) = –8 N (opposition)

Mass (m) = 100 Kg

<h3>Deceleration (a) =? </h3>

<h3>F = ma</h3>

–8 = 100 × a

Divide both side by 1000

a = \frac{-8}{100}

<h3>a = –0.08 ms¯²</h3>

Therefore, the deceleration of the box is –0.08 ms¯²

Finally, we shall determine the time taken for the box to come to rest. This can be obtained as follow:

Deceleration (a) = –0.08 ms¯²

Initial velocity (u) = 21 ms¯¹

Final velocity (v) = 0 ms¯¹

<h3>Time (t) =.? </h3>

<h3>v = u + at</h3>

0 = 21 + (–0.08×t)

0 = 21 – 0.08t

Collect like terms

0 – 21 = –0.08t

–21 = –0.08t

Divide both side by –0.08

t = \frac{-21}{-0.08}\\\\

<h3>t = 262.5 s</h3>

Therefore, it will take 262.5 s for the box to come to rest.

Learn more: brainly.com/question/14446351

7 0
2 years ago
Real bodies emit and absorb more radiation than a blackbody at the same temperature. True or False
monitta

Answer: False

Explanation:

Relative to the concept of radiations, a black body is an object capable of absorbing any form of electromagnetic radiation irrespective of its frequency or angle of incidence when incident on such object.

However, the same cannot be said about real bodies as real bodies are those which reflect all rays incident on them completely and uniformly in all directions.

One very important characteristic of black bodies is that they are ideal emmiters.

The concept of emmisivity is brought about by the existence of real bodies .

This is due to the fact that they are only able to emit radiation at a fraction of the black body energy levels.

Please note that by convention, the emmisivity of a real body is always less thaan 1.

As such they are not able to emit as much radiation as a black body at the same temperature.

3 0
3 years ago
The gases cast off by a dying star are called
lozanna [386]
Planetary Nebulae hope that helps
4 0
3 years ago
Read 2 more answers
Question 1
irina [24]
Dang that’s a lot wait i will answer in comments
5 0
2 years ago
Read 2 more answers
Other questions:
  • A stone is thrown vertically upward with a speed of 18.0 . (a)How fast is it moving when it reaches a height of 11.0 ? (b)How lo
    6·1 answer
  • Explain why the car reaches a top speed even though the thrust Force remains constant at 3500N
    12·1 answer
  • A truck covers 40.0 m in 7.50 s while uniformly slowing down to a final velocity of 2.55 m/s. (a) Find the truck's original spee
    14·1 answer
  • Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit.
    9·1 answer
  • In what state must matter exist for fusion reactions to take place?<br><br> ...?
    7·2 answers
  • Explain why elements in the same family have similar physical and chemical properties
    11·1 answer
  • What property of most stars can be determined by their color?
    13·1 answer
  • What are power tools​
    15·2 answers
  • What is the force of a 100 kg car that hits a wall while accelerating at 5 m/s2?​
    5·1 answer
  • A 3kg plastic tank that has a volume of 0.2m ^3 Is filled with liquid water . Assuming the density of water is 1000kg/m^3 determ
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!