Sound intensity is measured on a logarithmic scale

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The discussion focuses on two main problems: calculating the force per unit area from wind and understanding sound intensity on a logarithmic scale. For the wind problem, participants are asked to derive a formula for wind force and determine the power output required for a cyclist facing various headwinds. The sound intensity problem involves explaining why sound intensity is logarithmic, calculating the intensity of sound near a lightning strike, and determining how intensity changes with distance. Participants also explore the reasons thunder is rarely heard beyond 10 miles. The conversation emphasizes the need for formulas and calculations related to both wind force and sound intensity.
matt23
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Help please!

Here is the prob and i don't not know where or how to begin.

1.Jose was having a really bad day out on the bike. He should have checked the forecast. First, the wind coming out of nowhere. Must have been a steady 30 mph with gusts up to 50, head on. Then the lightning. First 1 km away, then 100 m away, then just 10 meters. Scary. Then the hail. First the size of a pea, then the size of a golf ball, then the size of a baseball!

Find a formula for the amount of force per unit area coming from wind blowing at a speed of S miles
per hour. As a result, supposing that a cyclist has an exposed area of 0:5 meter2, nd how much power
(in watts) the cyclist must put out to maintain a speed of 10 mph into a headwind of 10, 20, and 30 miles per hour.The top professional cyclists can produce up to a kilowatt of power in short bursts. What would this equate to in wind speed?


And the second prob is:
Sound intensity is measured on a logarithmic scale. Explain why this is the case. Find a gure for the sound intensity (in watts per centimeter2) for a person in close proximity to a lightning strike. Then give a formula for the sound intensity of the same strike when the person is R meters from the lighting strike. If a lightning bolt produces thunder at 130 dB at 10 meters, what is the intensity at 100 m? At 1 km? Can you explain why thunder is rarely heard at a distance more than 10 miles?
 
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matt23 said:
Here is the prob and i don't not know where or how to begin.

1.Jose was having a really bad day out on the bike. He should have checked the forecast. First, the wind coming out of nowhere. Must have been a steady 30 mph with gusts up to 50, head on. Then the lightning. First 1 km away, then 100 m away, then just 10 meters. Scary. Then the hail. First the size of a pea, then the size of a golf ball, then the size of a baseball!

Find a formula for the amount of force per unit area coming from wind blowing at a speed of S miles
per hour. As a result, supposing that a cyclist has an exposed area of 0:5 meter2, nd how much power
(in watts) the cyclist must put out to maintain a speed of 10 mph into a headwind of 10, 20, and 30 miles per hour.The top professional cyclists can produce up to a kilowatt of power in short bursts. What would this equate to in wind speed?

What is the equation for the force of air drag?

And the second prob is:
Sound intensity is measured on a logarithmic scale. Explain why this is the case. Find a gure for the sound intensity (in watts per centimeter2) for a person in close proximity to a lightning strike. Then give a formula for the sound intensity of the same strike when the person is R meters from the lighting strike. If a lightning bolt produces thunder at 130 dB at 10 meters, what is the intensity at 100 m? At 1 km? Can you explain why thunder is rarely heard at a distance more than 10 miles?

What is the equation for sound intensity?
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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