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SVETLANKA909090 [29]
3 years ago
8

A plate in a parallel-plate capacitor has an area of 0.03 m2 and is 0.5 × 10–3 m from the other plate. The space between the pla

tes is filled with a dielectric that has a dielectric constant of 6.7. Determine the capacitance of the capacitor. Recall that ε0 = 8.85 × 10–12 Startfraction coulombs squared over Newtons and meters squared Endfraction.
Physics
1 answer:
aivan3 [116]3 years ago
7 0

Answer:

4 x 10^-9F

Explanation:

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Please help with this!!!!!
34kurt
(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K)  I hoped that helped
4 0
3 years ago
Who was the first scientist to question the idea that atoms were uncuttable
gizmo_the_mogwai [7]

Answer:

Democritus

Explanation:

He called these "uncuttable" pieces atomos. This is where the modern term atom comes from. Democritus first introduced the idea of the atom almost 2500 years ago.

5 0
3 years ago
The spectra of most stars are dark-line spectra because ________.
pav-90 [236]
Because dark line spectra result from passing white light through ionized gasses and plasmas, which is what the atmosphere of stars are made of.  These frequencies are scattered by the star's atmosphere as it leaves the surface (photosphere) of the star, and don't make it to earth.
5 0
3 years ago
A negative charge, q1, of 6 µC is 0. 002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the e
prohojiy [21]

Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

<h3>What is electrical force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The given data in the problem is

q₁ is the negative charge = 6 µC=6×10⁻⁶ C

q₂ is the positive charge = 3 µC=3×10⁻⁶ C

r is the distance between the charges=0.002 m

F_E is the electric force =?

The value of electric force will be;

\rm F_E= \frac{Kq_1q_2}{r^2} \\\\ F_E= \frac{9\times 10^9\times 6\times 10^{-6}\times3\times10^{-6}}{(0.002)^2}\\\\ \rm F_E=4.05\times10^4\;N

Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

To learn more about the electrical force refer to the link;

brainly.com/question/1076352

7 0
2 years ago
The emission of x rays can be described as an inverse photoelectric effect.
timama [110]

Answer:

The potential difference through which an electron accelerates to produce x rays is 1.24\times 10^4\ volts.                                                  

Explanation:

It is given that,

Wavelength of the x -rays, \lambda=0.1\ nm=0.1\times 10^{-9}\ m

The energy of the x- rays is given by :

E=\dfrac{hc}{\lambda}

The energy of an electron in terms of potential difference is given by :

E=eV

So,

\dfrac{hc}{\lambda}=eV

V is the potential difference

e is the charge on electron

V=\dfrac{hc}{e\lambda}

V=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-19}\times 0.1\times 10^{-9}}

V = 12431.25 volts

or

V=1.24\times 10^4\ volts

So, the potential difference through which an electron accelerates to produce x rays is 1.24\times 10^4\ volts. hence, this is the required solution.

4 0
3 years ago
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