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zaharov [31]
2 years ago
4

Two parallel plate capacitors are identical, except that one of them is empty and the other contains a material with a dielectri

c constant of 4.2 in the space between the plates. The empty capacitor is connected between the terminals of an ac generator that has a fixed frequency and rms voltage. The generator delivers a current of 0.29 A. What current does the generator deliver after the other capacitor is connected in parallel with the first one?
Physics
1 answer:
weeeeeb [17]2 years ago
6 0

Answer:

current = 1.51 A

Explanation:

Initially the capacitor without any dielectric is connected across AC source

so the capacitive reactance of that capacitor is given as

x_c = \frac{1}{\omega c}

now we have

i = \frac{V_{rms}}{x_c}

here we know that

i = 0.29 A

now other capacitor with dielectric of 4.2 is connected in parallel with the first capacitor

so here net capacitance is given as

c_{eq} = 4.2c + c = 5.2c

now the equivalent capacitive reactance is given as

x_c' = \frac{1}{\omega(5.2c)}

x_c' = \frac{x_c}{5.2}

so here we have new current in that circuit is given as

i' = \frac{V_{rms}}{x_c'}

i' = 5.2 (i) = 5.2(0.29)

i' = 1.51 A

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Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm sli
Orlov [11]

Answer:

a.3.20m

b.0.45cm

Explanation:

a. Equation for minima is defined as: sin \theta=\frac{m\lambda}{\alpha}

Given m=3,\lambda=6.33\times 10^-^7 and \alpha=0.00015:

#Substitute our variable values in the minima equation to obtain \theta:

\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad

#draw a triangle to find the relationship between \theta, y \ and L.

tan(\theta)=y/L               #where y=4.05cm

L=y/tan(\theta)=3.20

Hence the screen is 3.20m from the split.

b.  To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

sin \theta=\frac{m\lambda}{\alpha}, where m=4 given that it's the minima with the smallest distance.

sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad

#we then use tan(\theta)=y/L to calculate L=4.5cm

Then from the equation subtract y_3 from y:

4.50cm-4.05cm=0.45cm

Hence, the distance \bigtriangleup y is 0.45cm

8 0
2 years ago
A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it
Degger [83]

Answer:

a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

d)\alpha_2=4.1454\,rad.s^{-2}

e)t_2=1.061\,s

Explanation:

Given that:

  • initial speed of turntable, N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

4 0
2 years ago
A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float
storchak [24]

Answer:

 t ’= \frac{1450}{0.6499 + 2 v_r},  v_r = 1 m/s       t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

        v_b - v_r = d / t

By the time the boat goes down the river

        v_b + v_r = d '/ t'

let's subtract the equations

       2 v_r = d ’/ t’ - d / t

       d ’/ t’ = 2v_r + d / t

       t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }

In the exercise they tell us

         d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

         d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

      t ’= \frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \  + 2 \ v_r}

      t ’= \frac{1450}{0.6499 + 2 v_r}

In order to complete the calculation, we must assume a river speed

          v_r = 1 m / s

       

let's calculate

      t ’= \frac{ 1450}{ 0.6499 + 2 \ 1}

      t ’= 547.19 s

8 0
2 years ago
Alculate the potential difference if 20J of energy are transferred by 8C of charge.
sveta [45]

Answer:

V = 2.5 J/C

Explanation:

<u><em>Given:</em></u>

Energy = E = 20 J

Charge = Q = 8 C

<u><em>Required:</em></u>

Potential Difference = V = ?

<u><em>Formula:</em></u>

V = \frac{E}{Q}

<u><em>Solution:</em></u>

V = 20/8

V = 2.5 J/C

6 0
2 years ago
A wheel completes 5.6 revolutions in 8 seconds.
nevsk [136]

Answer:

86.15\pi rad/min

Explanation: Angular velocity is the number of revolutions made per unit time.

We convert the number of revolutions to radians and the time given in seconds to minutes,

Given;

1rev=2\pi rad\\therefore\\5.6rev=5.6*2\pi rad\\= 11.2\pi rad

Also,

60s = 1 min

hence

8s=\frac{8}{60}min\\=0.13min

We now divide the number of revolution in radians by the time in minutes.

\omega =\frac{11.2\pi}{0.13min}\\\omega=86.15\pi rad/min

5 0
2 years ago
Read 2 more answers
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