by how many order of magnitute is the speed of light greater than the average speed of grey sound? Rose Diamond when he set the hight gate track 710 m record of for this 47.43 sec in April 2003
710m/47.43s = close to 15 m/s.
The speed of light is 3 x 10^8 m/s or 30 x 10^7 m/s. That is 30 x 10^7/15 = 20,000,000 give or take a little. That is ABOUT 7 orders of magnitude (count the zeros) faster for the speed of light over the track star. You don't mention how accurately you want this calculated.
thanks
To calculate the number of orders of magnitude by which the speed of light is greater than the average speed of sound, you need to compare the two speeds and determine the ratio between them.
The average speed of sound in dry air at room temperature is approximately 343 meters per second (m/s). We can calculate the ratio as follows:
Speed of light / Average speed of sound = (3 x 10^8 m/s) / (343 m/s)
To simplify this calculation, we can rewrite the average speed of sound as 3.43 x 10^2 m/s. Now the calculation looks like this:
(3 x 10^8 m/s) / (3.43 x 10^2 m/s)
To divide these two numbers, we divide their magnitudes and subtract the exponents:
(3 / 3.43) x 10^(8 - 2)
Which simplifies to:
(0.8746355685) x 10^6
Since we are looking for the order of magnitude, we round 0.8746355685 to 1. Therefore, the speed of light is approximately 10^6 times greater than the average speed of sound.
In other words, the speed of light is about 1 million times faster than the average speed of sound, so it is approximately 6 orders of magnitude faster.