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sergiy2304 [10]
3 years ago
9

Speed of light in air is 3 x 10 m/s, determine the speed of light in the glass fibre.

Physics
1 answer:
Dafna11 [192]3 years ago
4 0
30 speed of light in the glass
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Estimate the constant rate of withdrawal (in m3 /s) from a 1375 ha reservoir in a month of 30 days during which the reservoir le
kap26 [50]

Answer:

Explanation:

1 ha = 10⁴ m²

1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²

In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³

Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m

Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³

Let Q be the withdrawal in m³

Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶

Q = 26.20 x 10⁶ m³

rate of withdrawal per second

= 26.20 x 10⁶ / 30 x 24 x 60 x 60

= 26.20 x 10⁶ / 2.592 x 10⁶

= 10.11 m³ / s

6 0
3 years ago
Is V directly proportional to I (From V=IR)??<br>Or is V inversely proportional to I (From P=IV) ???
ziro4ka [17]

First of all, I is proportional V according to the Ohm's Law. R is merely a constant you need to obtain an equation. However, it is true that R changes with temperature and pressure, therefore Ohm's Law is only applicable in an invariable environment. Also this constant R is different for different materials.

So, do not get confused.

Ohm's law is not a universal law, please remember that as well. Some materials do not follow it and we call them non-ohmic conductors. I hope I helped! ^-^

3 0
3 years ago
Which of the following is produced as a result of photosynthesis
kenny6666 [7]
I think the answer is D but i could be wrong
3 0
3 years ago
Read 2 more answers
Why don't we see other stars during day?
Monica [59]

Answer:

Stars aren't visible during the sunlit hours of daytime because the light-scattering properties of our atmosphere spread sunlight across the sky. Seeing the dim light of a distant star in the blanket of photons from our Sun becomes as difficult as spotting a single snowflake in a blizzard.

Explanation:

5 0
3 years ago
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A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill acc
krek1111 [17]

Answer:

a

    \alpha  =  2327.7 \ rev/s^2

b

   \theta = 9124.5 \ rev

Explanation:

From the question we are told that

    The maximum  angular   speed is  w_{max} =  391000 \ rpm  =  \frac{2 \pi  *  391000}{60} =  40950.73 \ rad/s

     The  time  taken is  t =  2.8 \ s

     The  minimum angular speed is  w_{min}= 0 \ rad/s this is because it started from rest

     

Apply the first equation of motion to solve for acceleration we have that

       w_{max} =  w_{mini} +   \alpha  *  t

=>     \alpha  = \frac{ w_{max}}{t}

substituting values

       \alpha  = \frac{40950.73}{2.8}

       \alpha  =  14625 .3 \  rad/s^2

converting to rev/s^2

  We have

           \alpha  =  14625 .3 *  0.159155 \ rev/s^2

           \alpha  =  2327.7 \ rev/s^2

According to the first equation of motion the angular displacement is  mathematically represented as

       \theta  =  w_{min} * t  + \frac{1}{2} *  \alpha  *  t^2

substituting values

      \theta  =  0  * 2.8  + 0.5  *  14625.3  *  2.8^2

      \theta  = 57331.2 \  radian

converting to revolutions  

        revolution = 57331.2 *  0.159155

        \theta = 9124.5 \ rev

6 0
3 years ago
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