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3 years ago

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A power source supplies an electric potential difference of 9.0 V to the plates of a capacitor. The energy stored in the capacit

or is 9.8 × 10–4 J. The dielectric constant of the dielectric in the capacitor is 5.1.
The ratio of the area of the plates to the distance between the plates, rounded to two significant figures, is
____ × 10^5 m.

Quizlet says 5.4 as the answer. Not sure if that's right.

1 answer:

Anarel [89]3 years ago

6 0

Answer:

5.4

Explanation:

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Air is being pumped into a spherical balloon so that its volume increases at a rate of 150 cm3/s. How fast is the radius of the

Slav-nsk [51]

0.119cm/s is the radius of the balloon increasing when the diameter is 20 cm.

<h3>How big is a circle's radius?</h3>

The radius of a circle is the distance a circle's center from any point along its circumference. Usually, "R" or "r" is used to indicate it.

A circle's diameter cuts through the center and extends from edge to edge, in contrast to a circle's radius, which extends from center to edge. Essentially, a circle is divided in half by its diameter.

dv/dt = 150cm³/s

d = 2r = 20cm

r = 10cm

find dr/dt

Given that the volume of a sphere is calculated using

v = 4/3πr³

Consider both sides of a derivative

d/dt(v) = d/dt( 4/3πr³)

dv/dt = 4/3π(3r²)dr/dt = 4πr²dr/dt

Hence,

dr/dt = 1/4πr².dv/dt

dr/dt = 1/4π×(10)²×150

dr/dt = 1/4π×100×150

dr/dt = 0.119cm/s.

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A particle moves according to a law of motions=f(t),t≥0 Wheretismeasured in seconds andsin feet.a) Find the velocity at timet.b)

Ray Of Light [21]

Answer:

a) $v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}$

b) 3s

c) t < 3s

d) 1.8ft

e) particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

Explanation:

a)Suppose the equation for motion is:

$f(t) = \frac{9t}{t^2+9}$

Then the velocity is the derivative of the motion function

$v(t) = \frac{df(t)}{dt} = (9t(t^2+9)^{-1})^'$

From here we can apply product rule

$v(t) = (9t)^{'}(t^2+9)^{-1} + (9t)((t^2+9)^{-1})^'$

$v(t) = \frac{9}{t^2+9} - \frac{9t(2t)}{(t^2+9)^2}$

$v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}$

b) The particle is at rest when v(t) = 0:

$\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} = 0$

$\frac{9}{t^2+9} = \frac{18t^2}{(t^2+9)^2}$

Multiply 2 sides by $(t^2+9)^2$ we have:

$t^2 + 9 = 2t^2$

$t^2 = 9$

$t = 3s$

(c) The particle is moving in positive direction when v(t) > 0:

$\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} > 0$

$\frac{9}{t^2+9} > \frac{18t^2}{(t^2+9)^2}$

Multiply 2 sides by $(t^2+9)^2$ we have:

$t^2 + 9 > 2t^2$

$t^2 < 9$

$t < 3s$

(d) As particle is moving in positive direction when t < 3s and negative direction when t > 3s, we can calculate the distance it's moving up to 3s and then after 3s

$f(3) = \frac{9*3}{3^2+9} = \frac{27}{18} = 1.5 ft$

At 3s, particle is changing direction to negative, so its position at 6s is

$f(6) = \frac{9*6}{6^2+9} = \frac{54}{45} = 1.2 ft$

Therefore from 3s to 6s it would have moved a distance of 1.5 - 1.2 = 0.3 ft

Then the total distance it has moved in the first 6 s is 1.5 + 0.3 = 1.8 ft

e) Acceleration is the derivative of velocity function:

$a(t) = \frac{dv(t)}{dt} = (\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2})^'$

$a(t) = -\frac{9(2(t^2+9))(2t)}{(t^2+9)^2} - \frac{36t}{(t^2+9)^2} + \frac{18t^2(2(t^2+9))(2t)}{(t^2+9)^3}$

$a(t) = -\frac{36t}{t^2+9} - \frac{36t}{(t^2+9)^2} + \frac{72t^3}{(t^2+9)^2}$

$a(t) = -\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2}$

Particle is speeding up when a(t) > 0:

$-\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2} > 0$

$\frac{72t^3 - 36t}{(t^2+9)^2} > \frac{54t}{t^2+9}$

as $t \& (t^2 + 9)^2 \geq 0$ we can multiply/divide both sides by it:

$8t^2 - 4 > 6(t^2+9)$

$8t^2 > 6t^2 + 58$

$t^2 > 29$

$t > \sqrt(29) \approx 5.385 s$

so particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

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4 years ago