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3 years ago

11

An isolated parallel-plate capacitor has a surface charge density. If the space between the plates is filled with a material of

dielectric constant k = 3, find the induced surface charge density ind on the dielectric?

1 answer:

anygoal [31]3 years ago

3 0

Answer:

Explanation:

$2\sqrt{34}$

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A rocket in deep space has an empty mass of 220 kg and exhausts the hot gases of burned fuel at 2500 m/s. What mass of fuel is n

Scilla [17]

Answer:

Explanation:

Let fuel is released at the rate of dm / dt where m is mass of the fuel

thrust created on rocket

= d ( mv ) / dt

= v dm / dt

this is equal to force created on the rocket

= 220 dv / dt

so applying newton's law

v dm / dt = 220 dv / dt

v dm = 220 dv

dv / v = dm / 220

integrating on both sides

∫ dv / v = ∫ dm / 220

lnv = ( m₂ - m₁ ) / 220

ln4000 - ln 2500 = ( m₂ - m₁ ) / 220

( m₂ - m₁ ) = 220 x ( ln4000 - ln 2500 )

( m₂ - m₁ ) = 220 x ( 8.29 - 7.82 )

= 103.4 kg .

8 0

4 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

Two resistors, R1 and R2, are

dlinn [17]

The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:

1/(120.7 Ω) = 1/<em>R₁</em> + 1/(221.0 Ω)

1/<em>R₁</em> = 1/(120.7 Ω) - 1/(221.0 Ω)

<em>R₁</em> = 1 / (1/(120.7 Ω) - 1/(221.0 Ω)) ≈ 265.9 Ω

3 0

3 years ago

Calculate the total energy of 4.0 kg object moving horizontally at 20 m/s 50 meters above the surface.

Serhud [2]

Answer:

Correct answer: E total = 2,800 J

Explanation:

Given:

m = 4 kg the mass of the object

V = 20 m/s the speed (velocity) of the object

H = 50 m the height of the object above the surface

E total = ? J

The total energy of an object is equal to the sum of potential and kinetic energy

E total = Ep + Ek

Ep = m g H we take g = 10 m/s²

Ep = 4 · 10 · 50 = 2,000 J

Ek = m V² / 2

Ek = 4 · 20² / 2 = 2 · 400 = 800 J

E total = 2,000 + 800 = 2,800 J

E total = 2,800 J

God is with you!!!

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3 years ago

Sally takes two bar magnets and randomly tapes one end of each bar magnet. she labels the magnets A and B. She brings the taped

Artyom0805 [142]

I already answered this quesiton. The fact is that there are only two kind of poles and since the two taped poles of the magnets labeled A and B attracts one to each other, we know that the two taped poles of the first two magnets are oppsosite.

Then, the taped pole of the third magnet has to be equal to one of the first two taped poles and opposite to the other of the first two taped poles.

That drives you to conclude (predict) that when she brings the taped end of the third magnet (magnet C) near each of the first two magntes, in one case they will attract each other and in the other case they will repele mutually.

4 0

3 years ago