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jeka57 [31]
3 years ago
7

What is the time lapse between seeing a lightning strike and hearing the thunder if the lightning flash is 47 km away? The speed

of light waves in air is 3 × 108 m/s and the speed of sound waves in air is 333 m/s.
Physics
1 answer:
Natali5045456 [20]3 years ago
6 0

Answer:

141.14098 secs

Explanation:

Time taken to see the lightning flash can be gotten from:

Velocity = distance/time

Time = distance/velocity

Time = (47 * 1000)/(3 * 10^8)

Time = 0.0001567 secs

Time taken to hear the thunder can be gotten from:

Velocity = distance/time

Time = distance/velocity

Time = (47 * 1000)/(333)

Time = 141.14114 secs

The time lapse between the lightning flash and the thunder will be:

141.14114 - 0. 0001567

= 141.14098 secs

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Explanation:

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What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
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Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

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