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kenny6666 [7]
3 years ago
5

A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric fiel

d. If the field strength is 713.0 N/C find the electric Tiux through the surface A) 560 Nm2/C B) 620 N·m2/C C) 160 n N.m2/C D) 280 N.m2/C
Physics
1 answer:
nexus9112 [7]3 years ago
8 0

Answer:

The electric flux is 280\ \rm N.m^2/C

Explanation:

Given:

  • Radius of the disc R=0.50 m
  • Angle made by disk with the horizontal \theta=30^\circ
  • Magnitude of the electric Field E=713.0\ \rm N/C

The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows

\phi=\int E.dA

where

  • \phi is the total Electric Flux
  • E is the Electric Field
  • dA is the Area through which the electric flux is to be calculated.

Now according to question we have

=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C

Hence the electric flux is calculated.

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Answer:

1 cm⁻¹ =1.44K  1 ev = 1.16 10⁴ K

Explanation:

The relationship between temperature and thermal energy is

     E = K T

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The Planck equation is

          E = h f

Let's start the transformations

     c = f λ = f / ν        

     f = c ν

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     h c ν = K T

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Let's replace the constants

     h = 6.63 10⁻³⁴ J s

     c = 3 10⁸ m / s

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     v = 1 cm-1 (100 cm / 1 m) = 10² m-1

   

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     T =  1.44K

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     T = 148K

1 Rydberg = 1.097 10 7 m

As we saw at the beginning the λ=1 / v

     T = (h c / K) 1 /λ

     T = 1,441 10⁻²  1 / 1,097 10⁷

     T = 1.3 10⁻⁹ K

    E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J

    E = KT

    T = E/K

    T = 1.6 10⁻¹⁹ /1.38 10⁻²³

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3 years ago
Use a common denominator to find-2/3 + -4/5​
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Answer:

-22/15

Explanation:

the least common denominator is  15 so first you multiply -2/3 by 5 in both the numerator and denominator making it -10/15

Then you do the same to -4/5 except you multiply the numerator and denominator by 3 giving you -12/15

If you add -10/15+ -12/15 you get -22/15

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Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.
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Explanation:

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