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miskamm [114]
3 years ago
14

A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte

r one time constant? You are supposed to use a constant e can you please include what the value should be for this and all other constants.
Physics
1 answer:
ivanzaharov [21]3 years ago
5 0

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

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nikdorinn [45]

Answer:

a

 \frac{d \phi_{E}}{dt}  =1.1977 *10^{10} \  V\cdot m/s

b

 I = 0.106 \  A

Explanation:

From the question we are told that

  The current is  I =  0.106 \  A

   The length of one side of the square a = 4.60 \  cm = 0.046 \  m

    The separation between the plate is  d = 4.0 mm  = 0.004 \ m

Generally electric flux is mathematically represented as

       \phi_E = \frac{Q}{\epsilon_o}

differentiating both sides with respect to t is  

       \frac{d \phi_{E}}{dt}  = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}

=>     \frac{d \phi_{E}}{dt}  = \frac{1}{\epsilon_o} *I

Here \epsilon_o is the permitivity of free space with value  

        \epsilon _o  =  8.85*10^{-12} C/(V \cdot m)

=>   \frac{d \phi_{E}}{dt}  = \frac{0.106}{8.85*10^{-12}}

=>   \frac{d \phi_{E}}{dt}  =1.1977 *10^{10} \  V\cdot m/s

Generally the displacement current between the plates in A

    I = 8.85*10^{-12} * 1.1977 *10^{10}

=>  I = 0.106 \  A

 

3 0
2 years ago
Which statement describes the force between charged particles
Anuta_ua [19.1K]
The force (attractive if the charges are dissimilar, else repulsive) is along a line that connects the two particles.
3 0
3 years ago
A soccer player takes a cor- ner kick, lofting a stationary ball 35.0° above the horizon at 22.5 m/s. If the soccer ball has a m
liberstina [14]

Answer:

(a)

x=7.83 Kgm/s

y=5.48 Kgm/s

(b)

191.25 N

Explanation:

(a)

Change in momentum in x direction

P_{x}=mvcos\theta where m is mass and v is the velocity and \theta is the angle of kick  

Substituting m=0.425 Kg, v=22.5m/s and \theta=35^{o}

P_{x}=0.425*22.5*cos 35= 7.833141424

P_{x}=7.83 Kgm/s

Change in momentum in y direction

P_{y}=mvsin\theta

P_{y}=0.425*22.5*sin 35= 5.484824673

P_{y}=5.48 Kgm/s

(b)

Force exerted by the player

F=mv/t where t is time

Substituting t=5*10^{-2} s

F=(0.425*22.5)/0.05= 191.25 N

8 0
2 years ago
Laser light of wavelength lambda passes through a thin slit of width a and produces its first dark fringes at angles of +/- 30 d
alukav5142 [94]

Answer:

D)  θ₂= 36. 6º  

Explanation:

In this diffraction experiment it is described by the equation

               sin θ = m λ

The first dark strip occurs for m = 1 and since the angle is generally small we can approximate sine to the value of the angle

              θ₁ = λ/ a

This equation is valid for linear slits, in the case of a circular slit the problem must be solved in polar coordinates, so the equation changes slightly

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In the proposed exercise we start with a linear slit of width a, where tes1 = 30º and end with a circular slit of the same diameter

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Let's clear (Lam/a) of equalizing the two equations

             θ₁ = θ₂/ 1.22

             θ₂ = 1.22 θ₁

             θ₂ = 1.22 30

             θ₂= 36. 6º

When reviewing the correct results is D

5 0
3 years ago
What happens when the clapper of an electric doorbell move to strike the bell?
faust18 [17]

Answer: The clapper of an electric doorbell is an electromagnet. When the clapper of a doorbell strikes the bell, it opens an electric circuit. The electromagnet of an electric motor is connected to a permanent magnet. Only the shaft of an electric motor turns when current flows through the motor.

Explanation: hope this helped

5 0
3 years ago
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