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3 years ago

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A 3.0 µF capacitor is initially connected to a 6.0 V battery. Once the capacitor is fully charged the battery is removed and a 4

.0 Ω resistor is connected between the two terminals of the capacitor. Find the current flowing through the resistor 14.0 µsec after the capacitor begins to be discharged.

1 answer:

Otrada [13]3 years ago

3 0

Answer:

$i=-66.6\times10^{-6}\;\;Amp$

Explanation:

Given,

$C=3\mu F\\V=6.0 volt\\R=4\Omega\\t=14\mu sec\\$

Now, charge in capacitor

$q_o=CV\\q_o=3\times10^{-6}\times6\\q_o=18\times10^{-6}\;\;C$

For discharging the RC circuit,

$q=q_oe^{-t/\tau}$

Differentiate with respect to the <em>'t'</em>

$\frac{dq}{dt}=-q_o\frac{t}{RC}e^{t/RC}\\i=-q_o\frac{t}{RC}e^{t/RC}\\i=-18\times10^{-6}\frac{14\times10^{-6}}{4\times3\times10^{-6}}e^{\frac{14\times10^{-6}}{4\times3\times10^{-6}}}\\i=-66.6\times10^{-6}\;\;Amp$

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A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin

astraxan [27]

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

$v_{1x} = 406 mph\\v_{1y} = 0$

While the components of the velocity of the wind are

$v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph$

So the components of the resultant velocity of the jet are

$v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph$

And the new speed is the magnitude of the resultant velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph$

6 0

3 years ago

An initially uncharged air-filled capacitor is connected to a 2.63 V charging source. As a result, 6.27×10−5 C of charge is tran

user100 [1]

Answer:

Charge stored after insertion will be $22.257\times 10^{-5}C$

Explanation:

We have given potential difference V =2.63 V

$Q_0=$ charge stored by the capacitor when without dielectric = $6.27\times 10^{-5}C$

We know that $Q_0=C_0V$ , here $C_0$ is capacitance without dielectric

$6.27\times 10^{-27}=2.63\times C_0$

$C_0=2.384\times 10^{-5}F$

We have given

k = dielectric constant = 5.99

C = Capacitance with the dielectric = k $C_0$ = 3.55× $2.384\times 10^{-5}$ = $8.463\times 10^{-5}F$

Potential difference is due to the external charging source so it remains same

V = potential difference after insertion = 2.63 volts

New charge stored , $Q=CV=8.463\times 10^{-5}\times 2.63=22.257\times 10^{-5}C$

7 0

4 years ago

Steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m.

Amiraneli [1.4K]

Answer:

Vb= 5.42 m/s

Vc= 5.33 m/s

$a=-1125\ m/s^2$

Δ L= 0.43 mm

Explanation:

a) A to B

Initial velocity Va = 0 m/s

As we know that

$V^2=U^2+2aS$

$V_b^2=2\times 9.81\times 1.5$

Vb= 5.42 m/s

b)

C to D

Vd= 0 m/s

$0^2=V_c^2-2\times 9.81\times 1.45$

Vc= 5.33 m/s

c)

Here

Vb= 5.42 m/s

Vc= 5.33 m/s

We know that

Acceleration is the rate of change of velocity

$a=\dfrac{V_c-V_b}{t}$

$a=\dfrac{5.33-5.42}{8\times 10^{-5}}$

$a=-1125\ m/s^2$

d)

Now length compressed given as

$V_c^2=V_b^2+2a\Delta L$

$5.33^2=5.42^2-2\times 1125\times \Delta L$

Δ L= 0.43 mm

3 0

4 years ago

The author’s feeling about a subject or topic, which is evidenced in word choice, is called __________.

MatroZZZ [7]

The author’s feeling about a subject or topic, which is evidenced in word choice, is called tone. <span>They could be literal meanings which are meanings explicitly given to the readers or the person being told to. In this type of meaning, the person doesn’t need to think the words too much or ponder on the meaning because they could easily understand what the other person or the author implies. Meanwhile connotative meanings are meanings using different figures of speech or symbolism that the person still needs to think it over before knowing the actual meaning of the word, sentence or story.</span>

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4 years ago

Read 2 more answers

A certain dam generates 120 MJ of mechanical (hydroelectric) energy each minute. If the conversion from mechanical to electrical

Karolina [17]

Answer:

electric energy ( power ) = 300000 W

Explanation:

given data

mechanical (hydroelectric) energy = 120 MJ/min = 2000000 J/s

efficiency = 15 % = 0.15

solution

we know that Efficiency of electric engine is expression as

Efficiency = Mechanical energy ÷ electric energy ......................1

and here dam electrical power output is

put here value in equation 1

electric energy ( power ) = Efficiency × Mechanical energy ( power )

electric energy ( power ) = 0.15 × 2000000 J/s

electric energy ( power ) = 300000 W

8 0

3 years ago