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
lorasvet [3.4K]
3 years ago
7

Can I have help? I REALLY need it! PLEASE!

Physics
1 answer:
valentina_108 [34]3 years ago
7 0
Parallel .............
You might be interested in
Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha
Westkost [7]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, A_p = 180 cm^2

- The charge on each plate, q = 17 * 10^-^6 C

- Permittivity of free space, e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}

- The radius for the flux region, r = 3.3 cm

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

                             A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ = \frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

                         E+ = E- = \frac{sigma}{2*e_o} \\\\

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

                        E_n_e_t = (E+)  + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C}  \\

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C

5 0
3 years ago
As denizens of the surface of a spinning planet, we are always in uniform circular motion. Imagine you are in Nairobi (on the Ea
Klio2033 [76]

Answer:

6.02 radian

Explanation:

At 12 noon position is zero radian on Monday

at 12 midnight the position is π/2

at 11 am position on Tuesday is 2π/24 x 23 ( after 23 hours ) = 6..02 radian.

7 0
3 years ago
I can someone help me and also I gtg so ill be back in a bit
Korvikt [17]

Answer: No sure what to do

Explanation: points for points

3 0
3 years ago
A sled travels 15 meters down a slope inclined at 30 degrees with a horizontal. What is the horizontal displacement of the sled?
Arte-miy333 [17]

Length of the slope is given as

L = 15 m

also the inclination is given as

\theta = 30 degree

now the horizontal displacement is let say "x"

now from geometry we can say

\frac{x}{L} = cos30

x = L cos30

now substitute all values in it

x = 15 * cos30

x = 12.99 m

<em>so it will displace horizontally by 12.99 m</em>

6 0
3 years ago
Hi im a little stuck on this question that came from my textbook in class it got a little confusing for me because i really dont
koban [17]

Consult the attached free body diagram.

By Newton's second law, the net force on the crate acting parallel to the surface is

∑ F[para] = (370 N) cos(-20°) - f = 0

(this is the x-component of the resultant force)

where

• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force

• f = magnitude of kinetic friction

The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.

Solve for f :

f = (370 N) cos(-20°) ≈ 347.686 N

The net force acting perpendicular to the surface is

∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0

(this is the y-component of the resultant force)

where

• n = magnitude of normal force

• 1480 N = weight of the crate

• (370 N) sin(-20°) = magnitude of the vertical component of push

The crate doesn't move up or down, so it's also in equilibrium in this direction.

Solve for n :

n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N

Then the coefficient of kinetic friction is µ such that

f = µn   ⇒   µ = f/n ≈ 0.216

7 0
2 years ago
Other questions:
  • An airplane is initially flying horizontally (not gaining or losing altitude), and heading exactly North. Suppose that the earth
    5·1 answer
  • If a beam of light enters one medium from another medium, what quantity does not change?
    7·1 answer
  • A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when
    9·1 answer
  • A 3.1-kilogram gun initially at rest is free to
    11·2 answers
  • Coin-shaped compartment that contains light-absorbing molecules
    6·1 answer
  • AM radio frequencies range between 550 kHz (kilohertz) and 1600 kHz and travel at the same speed, 3.0 x 108 m/s. What is the wav
    8·2 answers
  • A situation known as ( blank)
    6·2 answers
  • I NEED HELP :) ty.............
    8·1 answer
  • If the voltage drop across the first resistor is 13.00V, then how many volts remain for the 2nd resistor?
    6·2 answers
  • What is the mass of an object if a force of 17 N causes it to accelerate at 1.5 m/s/s?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!