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drek231 [11]
3 years ago
5

The sine of the incident angle is 0.217; the sine of the refracted angle is 0.173. calculate the index of refraction.

Physics
2 answers:
notka56 [123]3 years ago
5 0
This can be solve using snell's law. snell's law equation is :

N1 / N2 = sin a2 / sin a1
where N1 is the index of  refraction of the air which is equal to 1
N2 is the index of refraction of the medium
a2 is the angle of refraction
a1 is the incident angle

subsitute the given values
1 / N2 = 0.173 / 0.217
N2 = 1 ( 0.217 / 0.173)
N2 = 1.25 is the index of refraction
Olenka [21]3 years ago
5 0
The sine of the incident angle is 0.217; the sine of the refracted angle is 0.173. calculate the index of refraction.<span>

n = 1.25

78% sure
</span>
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w is angular frequency

formula:2π/T

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F/R-->mw²

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What is the difference between real and apparent weightlessness?
MrRissso [65]

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4 0
2 years ago
It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i
timama [110]

Answer:

The mass of Uranium present in a 1.2mg sample is 4.8 \cdot 10^{-6}\,mg

Explanation:

The ration between Uranium mass and total sample mass is: \frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}

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7 0
3 years ago
When a piano tuner strikes both the A on the piano and a 440 Hz tuning fork, he hears a beat every 2 seconds. The frequency of t
AysviL [449]

Answer:

The  frequency is  f = 439.5 \  Hz        

Explanation:

From the question we are told that

   The frequency of the  tuning fork is  f_t  =  440 \ Hz

   The beat period is  T  =  2 \  s

Generally the beat frequency is mathematically represented as

       f_b  =  \frac{1}{T}

       f_b  =  \frac{1}{2}

      f_b  = 0.5 \  Hz

The  beat frequency is also represented mathematically as

     f_b  =  f_t  -  f

Where  f is the frequency of the piano

 So

       f =  440 -  0.5  

       f = 439.5 \  Hz        

3 0
3 years ago
A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece
seraphim [82]

Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

4 0
2 years ago
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