Ask question

Ask question

All categories

3 years ago

14

The capacitor is now reconnected to the battery, and the plate separation is restored to d. A dielectric plate is slowly moved i

nto the capacitor until the entire space between the plates is filled. Find the energy U2 of the dielectric-filled capacitor. The capacitor remains connected to the battery. The dielectric constant is K.Express your answer in terms of A, d, V, K, and ϵ0.

1 answer:

Paha777 [63]3 years ago

8 0

Answer:

U2 = KAε0V2 / (2d)

Explanation:

The dielectric constant K just replaces the "3″ from Part B.

You might be interested in

An astronaut is rotated in a horizontal centrifuge at a radius of 5.0 m. (a) What is the astronaut’s speed if the centripetal ac

sergeinik [125]

Explanation:

Given that,

Radius of circular path, r = 5 m

Centripetal acceleration, $a=7g\ m/s^2$

(a) Let v is the astronaut’s speed. The formula for the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$v=\sqrt{ar}$

$v=\sqrt{7\times 9.8\times 5}$

v = 18.5 m/s

(b) Let T denotes the time period. It is given by :

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi \times 5}{18.5}$

T = 1.69 s

Let N is the number of revolutions. So,

$N=\dfrac{60}{1.69}=35.5\ rev/min$

So, the number of revolutions per minute is 35.5

(c) T = 1.69 seconds

Hence, this is the required solution.

8 0

3 years ago

A car accelerates uniformly in a straight line

julia-pushkina [17]

The car travels a distance <em>d</em> from rest with acceleration <em>a</em> after time <em>t</em> of

<em>d</em> = 1/2 <em>a</em> <em>t</em>²

It covers 69 m with 2.8 m/s² acceleration, so that

69 m = 1/2 (2.8 m/s²) <em>t</em>²

<em>t</em>² = 2 (69 m) / (2.8 m/s²)

<em>t</em> ≈ 7.02 s

where we take the positive square root because we're talking about time *after* the car begins accelerating.

8 0

3 years ago

Juliette is driving her car when she sees a cat run across the road. If she is able to stop the car over a distance of 0.025km i

jek_recluse [69]

$velocity=\frac{distance}{time}=\frac{0,025km}{2,5s}=\frac{25m}{2,5s}=100\frac{m}{s}\\\\
acceleration=\frac{velocity}{time}=\frac{-10\frac{m}{s}}{2,5s}=-4\frac{m}{s^2}\\\\Solution\ is\ D.$

6 0

3 years ago

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

<u>For the course of upward acceleration:</u>

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

d)

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

2 years ago

Two airplanes leave an airport at the same time. The velocity of the first airplane is 750 m/h at a heading of 51.3°. The veloci

abruzzese [7]

Refer to the diagram shown below.

In 2.4 hours, the distance traveled by the first airplane heading a 51.3° at 750 mph is
a = 750*2.4 = 1800 miles.

The second airplane travels
b = 620*2.4 = 1488 mile

The angle between the two airplanes is
163° - 51.3° = 111.7°

Let c = the distance between the two airplanes after 2.4 hours.
From the Law of Cosines, obtain
c² = a² + b² - 2ab cos(111.7°)
= 3.24 x 10⁶ + 2.2141 x 10⁶
c = 2335.41 miles

Answer: 2335.4 miles

4 0

3 years ago

Read 2 more answers