Respuesta :
Answer:
Light takes less time than sound.
Explanation:
Let's say, the teacher and the student are at a distance "d" from each other.
The medium around them would be air.
And,
The speed of light in air is approx. 3× 10⁸ m/s
while, the speed of sound in air is approx. 330 m/s
We have a formula that establishes the relation between speed, distance and time.
[tex] \boxed{ \mathsf{speed = \frac{distance}{time} }}[/tex]
Our hunt for time — Speed in both the scenarios is known to us whereas the distance is same.
Sound
[tex] \mathsf{330 = \frac{d}{time_{s}} }[/tex]
[tex] \underline{\mathsf{time _{s} = \frac{d}{330} }}[/tex]
Light
[tex] \mathsf{3 \times {10}^{8} = \frac{d}{time _{l} } }[/tex]
[tex] \underline{ \mathsf{ time _{l} = \frac{d}{3 \times {10}^{8}} }}[/tex]
The best way of comparison is finding their ratio.
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ \frac{d}{330} }{ \frac{d}{3 \times {10}^{8} } } }[/tex]
simplifying the fraction
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{d \times (3 \times {10}^{8} )}{330 \times d}}[/tex]
d gets canceled and we're left with the following expression
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ (3 \times10 \times {10}^{7} )}{330}}[/tex]
30, being a common factor in the numerator as well as denominator, gets canceled out. and in its place remains 1/ 11
(why?
=> 30÷330 = 1÷11)
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ 1\times {10}^{7} }{11}}[/tex]
taking timeₛ to the numerator on the other side.
[tex] \implies \mathsf{time_{s} = \frac{ 1\times {10}^{7} }{11}\times time_{l}}[/tex]
Therefore, we get timeₛ is approx. 10⁶ times the timeₗ.
That's a big difference, no wonder light's way much faster than sound.
As lesser the time taken to cover a distance, faster is the wave.
The sound takes about 874,000 times MORE time than the light takes.
