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4vir4ik [10]
3 years ago
14

Evangelista Torricelli was the first person to realize that we live at the bottom of an ocean of air. He correctly surmised that

the pressure of our atmosphere is attributable to the weight of the air. The density of air at 0°C at the Earth's surface is 1.29 kg/m3. The density decreases with increasing altitude (as the atmosphere thins). On the other hand, if we assume that the density is constant at 1.29 kg/m3 up to some altitude h, and zero above that altitude, then h would represent the depth of the ocean of air.
(a) Use this model to determine the value of h that gives a pressure of 1.00 atm at the surface of the Earth.
(b) Would the peak of Mt. Everest rise above the surface of such an atmosphere?
Yes or no?
Physics
1 answer:
Radda [10]3 years ago
8 0

Answer:

a)    h = 8.02 10³ m  b) yes

Explanation:

a) The pressure in a fluid is given by

      P = ρ g h

The pressure in this case is the atmospheric pressure, 1.013 105 Pa, let's clear the height (h)

      h = P / ρ g

      h = 1.013 10⁵ / (1.29 9.8)

      h = 8.02 10³ m

b) The height of Mount Everest is 8848 m

It is above this height, according to this model there would be no air to breathe

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Answer:

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t_{pass} = \frac{23\ m}{9.84\ m/s} \\

<u>t_pass = 2.34 m</u>

<u></u>

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2as = v_{f}^{2} - v_{i}^2\\a = \frac{v_{f}^{2} - v_{i}^2}{2D}\\\\a = \frac{(0\ m/s)^{2}-(9.84\ m/s)^2}{(2)(23\ m)}\\\\a =  -2.1\ m/s^2

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