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3 years ago

Calculate the total resistance for ten 120 ohm resistors in series

Physics

1 answer:

GarryVolchara [31]3 years ago

7 0

The total effective resistance of several resistors in SERIES is just the sum of all the individual resistances.

So the effective resistance of ten 120-ohm resistors in series is

(120 + 120 + 120 + 120 + 120 + 120 + 120 + 120 + 120 + 120) .

A much easier way to write it is . . . (10) x (120) .

Total effective resistance = 1,200 ohms .

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Two narrow slits 0.02 mm apart are illuminated by light from a CuAr laser (λ = 633 nm) onto a screen. a) If the first fringe is

Masja [62]

Answer:

a) 0.063m

b) 2.72°

c) 3151 fringes

d) 1.87*10^-6m

Explanation:

a) To find the screen distance you use the following formula:

$y=\frac{m\lambda D}{d}\\\\D=\frac{dy}{m\lambda}$

D: screen distance

d: distance between slits

m: order of the fringes

λ: wavelength

By replacing the values of the parameters you obtain:

$D=\frac{(0.02*10^{-3}m)(0.2*10^{-2}m)}{(1)(633*10^{-9}m)}=0.063m$

b) The condition for dark fringes is given by:

$\lambda(m+\frac{1}{2})=dsin\theta$

for the first dark fringe the angle is:

$\theta=sin^{-1}(\frac{\lambda(m+\frac{1}{2})}{d})\\\\\theta=sin^{-1}(\frac{(633*10^{-9}m)(1+\frac{1}{2})}{0.02*10^{-3}m})=2.72\°$

c) the visible number of fringes is given by:

$N=1+2\frac{D}{d}=1+\frac{0.063m}{0.02*10^{-3}}=3151 \ fringes$

d) the wavelength of a laser in which its first order fringe coincides with the third one of the CuAr laser is:

$y=\frac{(3)(633*10^{-9}m)(0.063m)}{0.02*10^{-3}m}=5.98*10^{-3}m\approx0.59cm\\\\\lambda'=\frac{dy}{mD}=\frac{(0.02*10^{-3}m)(0.59*10^{-2}m)}{(1)(0.063m)}=1.87*10^{-6}m$

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3 years ago

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cestrela7 [59]

Answer:

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