A 3000-m-high mountain is located on the equator.
How much faster does a climber on top of the mountain move than a surfer at a nearby beach? The earth's radius is 6400 km.
Please use the numbers that are provided in the question so that I can see what I am doing incorrectly. Thank you very much in advance.
The angualr momentum speed(w) is the same,the linear speed(v)
V1=2(pi)r/t=>v=2(3.14)(6.4*10^6)/86400=465.18m/s (speed on the beach).
V2=2*3.14(6.4*10^6+3.0*10^3)/86400=465.40m/s (speed on the mountain top)
the speed difference=465.18-465.40
=0.22m/s
AND 86400 IS THE NUMBER OF SECONDS IN ONE DAY (24*60*60).
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