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3 years ago

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A 5.40 uF parallel-plate, air capacitor has a plate separation of 3.50 mm and is charged to a potential difference of 480 V. Cal

culate the energy density the region between the plates.

1 answer:

m_a_m_a [10]3 years ago

4 0

Answer:

Energy density = 0.0831 J/m³

Explanation:

Energy density of a capacitor is given by the expression

$u=\frac{1}{2}\epsilon E^2$

We have electric field ,

$E=\frac{V}{d}=\frac{480}{3.5\times 10^{-3}}=1.37\times 10^5V/m$

Here there is no dielectric

So energy density,

$u=\frac{1}{2}\epsilon_0 E^2=\frac{1}{2}\times 8.85\times 10^{-12} \times (1.37\times 10^5)^2=8.31\times 10^{-2}=0.0831 J/m^3$

Energy density = 0.0831 J/m³

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So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

tia_tia [17]

Answer:

a) $a=33.73mm/s^{2}$

b) mg>N

c) $\%_{change}=0.343\%$

d) $a=24.07mm/s^{2}$

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

$a_{c}=\omega ^{2}r$

where:

$\omega=\frac{2\pi}{T}$

we know the period of rotation of the earth is about 24 hours, so:

$T=24hr*\frac{3600s}{1hr}=86400s$

so we can now find the angular speed:

$\omega=\frac{2\pi}{86400s}$

$\omega=72.72x10^{-6} rad/s^{2}$

So the centripetal acceleration will be:

$a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)$

which yields:

$a_{c}=33.73mm/s^{2}$

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

$\sum F=0$

so we get that:

$N-mg+ma_{c} = 0$

and solve for the normal force:

$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$

and let's calculate the normal force:

$N=m(g-a_{c})$

$N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})$

N=586.58N

so now we can calculate the percentage change:

$\%_{change} = \frac{mg-N}{mg}x100\%$

so we get:

$\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%$

$\%_{change}=0.343\%$

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

$cos \theta = \frac{AS}{h}$

In this case:

$cos \theta = \frac{r}{R_{E}}$

so we can solve for r, so we get:

$r= R_{E}cos \theta$

in this case we'll use the average radius of earch which is 6,371 km, so we get:

$r = (6371x10^{3}m)cos (44.4^{o})$

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

$a=\omega ^{2}r$

$a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m$

$a=24.07 mm/s^{2}$

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3 years ago

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What's the difference between beam balance and spring balance?

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<span>A spring balance will give a different mass reading on the Moon from that on Earth. A beam balance however, will give the same reading.Explain why.

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<span>Beam Balance:

Beam balance measures mass.

Mass is the amount of matter the object has.

The S.I. unit of mass is kg

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Spring Balance:

Spring balance measures weight not mass.

Disadvantage: It requires gravity to measure.

The S.I. unit of weight is newton (N).

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Mass remains constant while weight is dependent on the gravitational pull of the planet. Mass only changes with a change in matter that results in a change of volume.

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4 years ago

A politician running for office tells a crowd at a rally that the only way to keep the food chain safe is to stop allowing the g

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Answer:

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3 years ago

A skydiver of mass m jumps from a hot air balloon and falls a distance d before reaching a terminal velocity of magnitude v. Ass

ahrayia [7]

Answer:

a) W = ΔK.E + ΔP.E = (mv²/2) - mgd

b) Power supplied by the drag force after the skydiver has reached terminal velocity = mgv

Explanation:

Using the work energy theorem, the work done by the drag force while the skydiver moves through a distance d is equal to the change in kinetic energy and potential energy of the body.

The potential energy of the skydiver drops by a magnitude of (mgd) falling through a distance of d.

ΔP.E = - mgd (skydiver loses this amount of potential energy)

But the skydiver falls from rest to gain a velocity of v after falling a distance of d vertically.

ΔK.E = (final kinetic energy at distance d) - (initial kinetic energy at rest)

Final kinetic energy at d = (1/2)(m)(v²) (the body is now falling with terminal velocity (v) at this point d)

Initial kinetic energy = 0 (since the skydiver was initially at rest)

ΔK.E = (1/2)(m)(v²) - 0 = (mv²/2)

W = ΔK.E + ΔP.E = (mv²/2) - mgd

b) At terminal velocity, the net force on the skydiver = 0

hence,

drag force = weight of the skydiver = mg

Power = F × v = mg × v = mgv

Hope this Helps!!!

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4 years ago