Post: Hardest math problem ever?
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04-24-2011, 12:32 AM
#1
iGir :D
Bounty hunter
For each breath that you take, how many of the air molecules would also been breathed by the patron saint of Physics, Sir Isaac Newton (1642-1727)during his life time, the atmosphere is about 8 km high, and the molecules in the air each occupy a space representing a little cubic box about 3.33*10^-9 m along a side, the earths radius is 6..38*10^6. make any reasonable assumptions for any data needed.
a) 6
b) 6*10^3
c) 6*10^6
d) 6*10^9
e) 6*10^12
Correct answer get's a present!
04-24-2011, 01:00 AM
#11
Solo
Rookie
Originally posted by iGir
Just a random math question.
For each breath that you take, how many of the air molecules would also been breathed by the patron saint of Physics, Sir Isaac Newton (1642-1727)during his life time, the atmosphere is about 8 km high, and the molecules in the air each occupy a space representing a little cubic box about 3.33*10^-9 m along a side, the earths radius is 6..38*10^6. make any reasonable assumptions for any data needed.
a) 6
b) 6*10^3
c) 6*10^6
d) 6*10^9
e) 6*10^12
Correct answer get's a present!
For each breath that you take, how many of the air molecules would also been breathed by the patron saint of Physics, Sir Isaac Newton (1642-1727)during his life time, the atmosphere is about 8 km high, and the molecules in the air each occupy a space representing a little cubic box about 3.33*10^-9 m along a side, the earths radius is 6..38*10^6. make any reasonable assumptions for any data needed.
a) 6
b) 6*10^3
c) 6*10^6
d) 6*10^9
e) 6*10^12
Correct answer get's a present!
Depends on how deep his breathes are. Can't be answered.
/thread :carling:
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04-24-2011, 01:02 AM
#12
DCLXVI
Smoke weed.
Originally posted by Dr.
The total volume of the atmosphere is 4 Re2 h, where Re is the Earth's radius, and h is the height of the atmosphere, giving 4.08 x 1018 m3. The fraction of air molecules ever breathed by the patron saint of Physics is thus 8.93 x 105 / 4.08 x 1018 = 2.19 x 10-13. The number of molecules in each breath of ours is the density 1/(3.3 x 10-9)3 = 2.78 x 1025 m-3, multiplied by the volume of each breath, 10-3 m3, or 2.78 x 1022 molecules. Multiplying by the fraction breathed by Newton, each breath of ours has about 6.08 x 109 molecules also breathed by him (D). Since we have about 9 x 108 breaths, each breath of ours has about 7 molecules also breathed by Isaac Newton.
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04-24-2011, 01:17 AM
#13
Tory Lanez
On bail right now
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04-24-2011, 01:23 AM
#16
Tory Lanez
On bail right now
04-24-2011, 01:28 AM
#18
Diezel
Banned
Assume 12 breaths per minute and a tidal volume of 1 liter.
= 12 liters/minute = .2 liters/second
Newton's age = 83
Assume avergage rate of intake between 0-20 = 1/2 of adult
Number of molecules Newton breathed
Volume inhaled lifetime = .2 l/s * (83-10)years
V = 4.6x10^8 liters
Volume of Earth's atmosphere = 4/3 π ((radius(earth+atm))^3 - (radius(earth))^3)
V.e.atm = 4/3 π ((6.378x10^6m)^3 - (6.370x10^6m)^3)
V.e.atm = 4 x10^18 m^3 = 4 x 10^21 liters
Number of molecules per breath
=.001m^3 (this is equal to 1 liter) / (3.33x10^-9)^3
= 2.7 x 10^22 molecules
Number that Newton breathed:
2.7x10^22 * 4.6x10^8 liters/4x10^21 liters
= 3.11 x10^9 molecules
What did I win????
= 12 liters/minute = .2 liters/second
Newton's age = 83
Assume avergage rate of intake between 0-20 = 1/2 of adult
Number of molecules Newton breathed
Volume inhaled lifetime = .2 l/s * (83-10)years
V = 4.6x10^8 liters
Volume of Earth's atmosphere = 4/3 π ((radius(earth+atm))^3 - (radius(earth))^3)
V.e.atm = 4/3 π ((6.378x10^6m)^3 - (6.370x10^6m)^3)
V.e.atm = 4 x10^18 m^3 = 4 x 10^21 liters
Number of molecules per breath
=.001m^3 (this is equal to 1 liter) / (3.33x10^-9)^3
= 2.7 x 10^22 molecules
Number that Newton breathed:
2.7x10^22 * 4.6x10^8 liters/4x10^21 liters
= 3.11 x10^9 molecules
What did I win????
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