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xz_007 [3.2K]
2 years ago
9

A parallel plate capacitor fully charged to voltage V is connected to the battery (the voltage on the plates remains fixed). If

the plates are pulled away from each other, how would it affect the charge on the capacitor
Physics
1 answer:
Papessa [141]2 years ago
7 0

Charge will decreases.

A parallel plate capacitor when it is fully charged to voltage V is given as:

                    C = Q/V

The capacitance of parallel plate capacitor with two plates of Area A separated by distance d and no dielectric material between plates is

                    C = ε₀ A /d

since from above equation it shows C is proportional to Q and also C is inversely proportional to distance d.

So, ATQ when d increases C will decrease which in result decreases charge on the capacitor.

Thus,  Charge will decrease.

Learn more about capacitance here:

     brainly.com/question/17115454

          #SPJ4

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Unpolarized light of intensity Io is incident on a stack of 7 polarizing filters, each with its axis rotated 17°cw with respect
mash [69]

The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

To answer the question, we need to know what polarization of light is.

<h3>What is polarization of light?</h3>

This is when the electric field vector of light is oscillating in one plane.

  • Now for light of intensity I' which is initially unpolarized, its intensity after polarization is I = 1/2I'.
  • Also, for light initially polarized, its intensity after polarization is I"' = I"cos²Ф where Ф is the angle between the initial direction and the direction of polarization.
<h3>Intensity of light through each polarized filter</h3>

Given that we have 7 polarizing filters, each rotated 17° cw with respect to the previous filter.

So, since the light is initially unpolarized,

  • The intensity through the first polarizing filter is I₁ = 1/2I₀ where I₀ is the initial intensity.
  • The intensity through the second polarizing filter is I₂ = I₁cos²17°= 1/2I₀cos²17°
  • The intensity through the third polarizing filter is I₃ = I₂cos²17° = 1/2I₀cos⁴17°
  • The intensity through the fourth polarizing filter is I₄ = I₃cos²17° = 1/2I₀cos⁶17°
  • The intensity through the fifth polarizing filter is I₅ = I₄cos²17° = 1/2I₀cos⁸17°
  • The intensity through the sixth polarizing filter is I₆ = I₅cos²17° = 1/2I₀cos¹⁰17°
  • The intensity through the seventh polarizing filter is I₇ = I₆cos²17° = 1/2I₀cos¹²17°.
<h3>The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity</h3>

Since I₇ is the last intensity I₇ = It = 1/2I₀cos¹²17°.

So, It/I₀ = 1/2cos¹²17°

= 1/2(0.9563)¹²

= 1/2 × 0.5850

= 0.2925

So, the ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

Learn more about intensity of polarized light here:

brainly.com/question/25402491

5 0
2 years ago
A ray diagram is shown.
Rufina [12.5K]

Answer:

X

Explanation:

8 0
3 years ago
Read 2 more answers
When a beam of charged particles moves through a magnetic field, what is the evidence that particles in the beam have momentum g
Vsevolod [243]

The charged particles are often deflated in a magnetic field.

<h3>What is a magnetic field?</h3>

The term charge refers to a positive or negative entity. The can be created when a charge is made to pass through a conductor in a magnetic field.

A magnetic field is created when we have a north pole and a south pole. The charged particles could be made to pass through the electric field and when that happens, we can see a pattern a shown in the image attached.

Thus, we can see that the charged particles are often deflated in a magnetic field.

Learn more about magnetic field:brainly.com/question/23096032

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4 0
1 year ago
A golf club hits a stationary 0.05kg golf ball with and average force of 5.0 x 10^3 newtons accelerating the ball at 44 meters p
maxonik [38]

Answer: The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

Explanation:

Force applied on the golf ball = 5.0\times 10^3 N

Mass of the ball = 0.05 kg

Velocity with which ball is accelerating = 44 m/s

Time period over which forece applied = t

f=ma=\frac{m\times v}{t}

t=\frac{0.05 kg\times 44m/s}{5.0\times 10^3 N}=4.4\times 10^{-4} seconds

Impulse=(force)\times (time)=f\times t = 5.0\times 10^3\times 4.4\times 10^{-4} seconds=2.2 Newton seconds

The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

7 0
3 years ago
You throw a glove straight upward to celebrate a victory. Its initial kinetic energy is K and it reaches a maximum height h. Wha
Rasek [7]

Answer:

K/2

Explanation:

The law of conservation of mechanical energy states that the sum of the kinetic and potential energies is a constant at any point.

At maximum height, the glove has purely potential energy but at the bottom, it has purely kinetic energy.

The potential energy at the top = kinetic energy at the bottom. The potential energy is given by

PE = mgh

At half height, this potential energy is

PE = \frac{1}{2}mgh

At this height, PE + KE = Constant = KE at bottom or PE at maximum height.

mgh = \frac{1}{2}mgh +KE

KE = \frac{1}{2}mgh = K/2

5 0
3 years ago
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