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A student witnesses a flash of lightning and then t=2.5s later the student hears assiciated clap of thunder. (please show work) (Part A) Sound travels at 343 m?s in the air, What distance from the student is the lightning strikes, in meters? (Part B) Light travels at 3.0x108 m/s in the air. How long, t1, in seconds did it take the light to reach the student's eyes after the flash?

Respuesta :

Answer:

857.5 m

2.8583×10⁻⁶ seconds

Explanation:

Time taken by the sound of the thunder to reach the student = 2.5 s

Speed of sound in air is 343 m/s

Speed of light is 3×10⁸ m/s

Distance travelled by the sound = Time taken by the sound × Speed of sound in air

⇒Distance travelled by the sound = 2.5×343 = 857.5 m

⇒Distance travelled by the sound = 857.5 m

Time taken by light = Distance the light travelled / Speed of light

[tex]\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow  \text{Time taken by light}=2.8583\times 10^{-6}[/tex]

Time taken by light = 2.8583×10⁻⁶ seconds

facundo3141592

We want to find the distance between the student and the lighting given that we know the time that passes between the moment he sees the lighting and the moment he hears it.

The solutions are:

  • a)  D = 857.501 m
  • b) t₁ = 2.858*10^(-6) s

Now, notice that there is an intrinsic problem with this question.

The student sees the "flash" when the light impacts on his eyes. Thus, the chain of events is:

  1. Lighting falls.
  2. Light travels and impacts the student's eyes, at this moment he sees the flash of light, this is what we consider the t = 0.
  3. 2.5 seconds after, sound arrives.

So there is a little time between when the lighting happens and the student sees it, that will also have an impact on the distance to the lighting.

then we have:

(3*10^8 m/s)*t₁ = D

(343m/s)*t₂ = D

Where t₁ is the time that the light needs to reach the student, t₂ is the time the sound needs to reach the student, and D is the distance between the student and the ligthing.

We also know that:

t₂ - t₁ = 2.5s

Then we can write:

t₂ = t₁ + 2.5s and replace it on the second equation:

(343m/s)*(t₁ + 2.5s) = D

Then we have a system of equations:

(3*10^8 m/s)*t₁ = D

(343m/s)*(t₁ + 2.5s) = D

We can see that both of these left sides are equal to the same thing, then we must have:

(3*10^8 m/s)*t₁ = (343m/s)*(t₁ + 2.5s)

Then we solve it for t₁

t₁*(3*10^8 m/s - 343m/s) =  (343m/s)*2.5s = 857.5m

t₁ = (857.5m)/(3*10^8 m/s - 343m/s) = 2.858*10^(-6) s

Now that we know the value of t₁, we can find the value of D.

(3*10^8 m/s)*t₁ = D

(3*10^8 m/s)* 2.858*10^(-6) s = D = 857.501 m

Then the solutions are:

  • a)  D = 857.501 m
  • b) t₁ = 2.858*10^(-6) s

If you want to learn more, you can read:

https://brainly.com/question/4598472

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