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Alex Ar [27]
3 years ago
5

he force between two long parallel conductors is 15 kg/metre. the conductor spacing is 10 cm. If one conductor carries twice the

current of the other, calculate the current in each conductor.

Physics
1 answer:
Oxana [17]3 years ago
8 0
Hey, I assumed you meant the force between conductors was 15 N/m not kg/m.  

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At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to (a) 2 times that of the Celsius and (b) 1/4 ti
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Answer:

(a) F = 320

(b) = F = -5.1625

Explanation:

The formula that converts degree Celsius (C) to degree Fahrenheit (F) is:

F = 1.8C + 32

Solving (a): F = 2C

Substitute 2C for F in the above equation

F = 1.8C + 32

2C = 1.8C + 32

Collect like terms

2C - 1.8C = 32

0.2C = 32

Multiply both sides by 5

5 * 0.2C = 32 * 5

C = 160

Recall that F = 2C

F = 2 * 160

F = 320

Solving (b): F = ¼C

Substitute ¼C for F in the above formula

F = 1.8C + 32

¼C = 1.8C + 32

Convert fraction to decimal

0.25C = 1.8C + 32

Collect like terms

0.25C - 1.8C = 32

-1.55C = 32

Divide both sides by -1.55

C = 32/(-1.55)

C = -32/1.55

C = -20.65

Recall that: F = ¼C

F = -¼ * 20.65

F = -5.1625

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In a chart the independent variable goes on the ___________. Question 5 options: left right middle
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A monochromatic light beam is incident on a barium target that has a work function of 2.50 \mathrm{eV} . If a potential differen
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The wavelength of the light beam required to turn back all the ejected electrons is 497 nm which is option (b).

  • Work function is a material property defined as the minimum amount of energy  required to infinitely remove electrons from the surface of a particular solid.
  • The potential difference required to support all emitted electrons is called the stopping potential which is given by v_0=\frac{K.E_m_a_x}{e} .....(1)
  • where v_0 is the stopping potential and e is the charge of the electron given by 1.6\times10^-^1^9 .

It is given that work function (Ф) of monochromatic light is 2.50 eV.

Einstein photoelectric equation  is given by:

K.E_m_a_x=E-\phi      ....(2)

where K.E(max) is the maximum kinetic energy.

Substituting (1) into (2) , we get

  ev_0=E-\phi\\1.6\times10^{-19} \times1=E-2.50\\E=1.6\times10^{-19}+2.50\\E=2.50eV

As we know that E=\frac{hc}{\lambda}  ....(3)

where Speed of light,c = 3\times10^8 m/s and Planck's constant , h = 6.63\times 10^-^1^9Js = 4.14\times 10^-^1^5 eVs

From equation (3) , we get

\lambda=\frac{hc}{E} \\\\\lambda=\frac{  4.14\times 10^-^1^5 \times 3 \times10^8}{2.50} \\\\\lambda=\frac{1240\times10^-^9}{2.50} \\\\\lambda=496.8\times10^-^9\\\\\lambda=497nm

Learn about more einstein photoelectric equation  here:

brainly.com/question/11683155

#SPJ4

8 0
1 year ago
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