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shtirl [24]
2 years ago
6

What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?

Physics
1 answer:
alukav5142 [94]2 years ago
5 0

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

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The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of
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Answer:

The angle between the blue beam and the red beam in the acrylic block is  

 \theta _d  =0.19 ^o

Explanation:

From the question we are told that

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     The  wavelength of the blue light is F  =  486.1 nm  =  486.1 *10^{-9} \ m

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    The incidence angle is  i  =  45^o

Generally from Snell's law the angle of refraction of the blue light  in the acrylic block  is mathematically represented as

       r_F =  sin ^{-1}[\frac{sin(i) *  n_a }{n_F} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_F =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.497} ]

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Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

       r_C =  sin ^{-1}[\frac{sin(i) *  n_a }{n_C} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_C =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.488} ]

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The angle between the blue beam and the red beam in the acrylic block

     \theta _d  =  r_C  - r_F

substituting values

       \theta _d  = 28.37 -  28.18

       \theta _d  =0.19 ^o

 

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If the force is balanced, the arrow is expect to continue in uniform motion but that is not the case as air resistance has massive impact on this body.

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