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4vir4ik [10]
2 years ago
11

There is an electric field in the region between the two plates. The magnitude of this electric field is ed. This imposes anothe

r condition on the charge densities on the surfaces of the plates. How can this condition be expressed?
Physics
1 answer:
krok68 [10]2 years ago
5 0

Answer:

it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

Explanation:

The configuration of parallel plates is called a capacitor and is widely used to create constant electric fields inside.

 To obtain this field it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

This is so that the fields created by each plate can be added inside and subtracted from the outside of the plates

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The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t , where the time tis in seconds. The
Reika [66]

Complete question:

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The

particle is momentarily at rest at t is:

Select one:

a. 9.3s

b. 1.3s

C. 0.75s

d.5.3s

e. 7.3s

​

Answer:

b. 1.3 s

Explanation:

Given;

position of the particle, x(t)=1 6t- 3.0t³

when the particle is at rest, the velocity is zero.

velocity = dx/dt

dx /dt = 16 - 9t²

16 - 9t² = 0

9t² = 16

t² = 16 /9

t = √(16 / 9)

t = 4/3

t = 1.3 s

Therefore, the particle is momentarily at rest at t = 1.3 s

6 0
2 years ago
Levi observed properties of four different waves and recorded observations about each one in his chart. A 2-column table with 4
makvit [3.9K]

Answer:

Wave W is a sound wave, Waves X and Y are light waves, and it is impossible to tell what kind of wave Wave Z is.

Explanation:

W      travels fastest through metal

X       travels fastest through air,

Y       travels more slowly through water than air    

Z       travels more slowly at cool temperatures

W appears to be sound wave  as sound travels fastest through metal .

X appears to be light wave as light travels fastest in air .

Y also appears to be light wave as  speed of light is reduced when it passes from air to water .

Z  It is impossible to tell anything about the nature of Z wave .

5 0
2 years ago
Read 2 more answers
Which of the following statements characterizing types of waves are true?
Tasya [4]

Answer:

a and b.

Explanation:

In general  types of wave

1. Transverse wave  :

    In these waves particle are vibrate perpendicular to motion of waves.

 Ex : Electromagnetic wave , Radio wave .

2. Longitudinal wave :

   In these waves particle are vibrate along the motion of waves.

 Ex : Sound wave

Mechanical wave  :

1 .These are transverse wave or Longitudinal wave or combination of them .

2.These waves required medium for propagation.

3. The particle are vibrate perpendicular to motion of waves.

So the option a and b are correct.

7 0
2 years ago
A square conducting loop 8.4 cm on a side is placed in a uniform B-field so that the plane of the loop is perpendicular to the d
arsen [322]

Answer:

Explanation:

area of square loop A = side²

= 8.4² x 10⁻⁴

A = 70.56 x 10⁻⁴ m²

when it is converted into rectangle , length = 14.7  , width = 2.1

area = length x width

= 14.7 x 2.1 x 10⁻⁴

= 30.87 x 10⁻⁴ m²

Let magnetic field be B

Change in flux = magnetic field x change in area

= B x ( 70.56 x 10⁻⁴ - 30.87 x 10⁻⁴ )

= 39.69 x 10⁻⁴ B

rate of change of flux = change in flux / time taken

= 39.69 x 10⁻⁴ B  / 6.5 x 10⁻³

= 6.1 x 10⁻¹ B

emf induced = 6.1 x 10⁻¹ B

6.1 x 10⁻¹ B  = 14.7 ( given )

B = 2.41 x 10

= 24.1 T

B ) magnetic flux is decreasing , so it needs to be increased as per Lenz's law . Hence current induced will be anticlockwise so that additional  magnetic flux is induced out of the page.

4 0
2 years ago
You are provided with three polarizers with filters making angles of (A) 90 ​∘ ​​ , (B) 180 ​∘ ​​ and (C) −45 ​∘ ​​ with respect
irinina [24]

Answer:

Order of maximum transmission of the polarizer is A, C and B.

Solution:

As per the question:

For the first polarizer, the angle is quite insignificant:

(A) 90^{\circ}:

The light intensity after passing through the first polarizer is I_{o} and this intensity does not depend on the angle of the polarizer.

Consider 90^{\circ} with the vertical, the intensity is given by:

I = I_{o}cos^{2}90^{\circ}

I = I_{o}cos(2(45^{\circ})) = I_{o}(\frac{1+cos90^{\circ})}{2} = \frac{I_{o}}{2}

(B) 180^{\circ}:

Suppose the second polarizer is  45^{\circ} with the vertical.

Now, intensity through the second polarizer:

I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(- 45 - 90)

I' =  \frac{I_{o}}{2}cos^{2}135^{\circ} = \frac{I_{o}}{4}

Now, if we consider the second polarizer to be 180^{\circ},

I' = \frac{I_{o}}{2}cos^{2}180^{\circ} = \frac{I_{o}}{2}cos^{2}(180^{\circ} - 90^{\circ}) = 0

(C) - 45^{\circ}:

Now,

Intensity through the third polarizer, if it is 180^{\circ} with the vertical:

I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(180 - (- 45))

I' = \frac{I_{o}}{8}

5 0
2 years ago
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