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Chapter 12: Problem 7

Atmospheric pressure is about \(10^{5} \mathrm{~N} / \mathrm{m}^{2}\), meaning that a \(100,000 \mathrm{~N}\) weight of air-corresponding to a mass of \(10,000 \mathrm{~kg}\) -sits atop very square meter of the ground (at or near sea level). If the air density were constant at \(1.25 \mathrm{~kg} / \mathrm{m}^{3}\) - rather than decreasing with height as it actually does - how high would the atmosphere extend to result in this weight (mass)?

Short Answer

Expert verified
Answer: The atmosphere would extend to 8000 meters.

Step by step solution

01

Calculate the force

We are given the atmospheric pressure (P) as \(10^5 N/m^2\) and the area (A) as \(1m^2\). We can find the force by using the formula: Force (F) = Pressure (P) * Area (A) F = \((10^5 N/m^2) * (1 m^2) = 10^5 N\)
02

Calculate the mass

Now that we have the force (F), we can find the mass by using the formula: Mass (m) = Force (F) / Gravity (g) Given g is approximately \(9.81m/s^2\), thus: m = \((10^5 N) / (9.81 m/s^2) ≈ 10^{4} kg\)
03

Calculate the height

We have found the mass (m) in a specific area. Now, we need to find the height (h) of the atmosphere. We are given the constant air density as \(1.25 kg/m^3\). We can find the height using the formula: Height (h) = Mass (m) / (Area (A) * Air Density) h = \((10^{4} kg) / (1 m^2 * 1.25 kg/m^3) = 8000 m\) Therefore, the height of the atmosphere with a constant air density of \(1.25 kg/m^3\) would extend to 8000 meters.

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Most popular questions from this chapter

The largest wind turbines have rotor diameters \(^{40}\) around \(150 \mathrm{~m}\). Using a sensible efficiency of \(50 \%\), what power does such a jumbo turbine deliver at a maximum design wind speed of \(13 \mathrm{~m} / \mathrm{s} ?\)

A hard slap might consist of about \(1 \mathrm{~kg}\) of mass moving at \(10 \mathrm{~m} / \mathrm{s}\). How much kinetic energy is this, and how much warmer would \(10 \mathrm{~g}\) of \(\mathrm{skin}^{33}\) get if the skin has the heat capacity properties of water, as in the definition of a calorie (Sec. 5.5; p. 73 and Sec. \(6.2\); p. 85 are relevant)?

Referring to Figure 12.7, examine performance at \(5 \mathrm{~m} / \mathrm{s}\) and at \(10 \mathrm{~m} / \mathrm{s}\), picking a representative power for each in the middle of the cluster of black points, and assigning a power value from the left-hand axis. What is the ratio of power values you read off the plot, and how does this compare to theoretical expectations for the ratio going like the cube of velocity?

Thermal energy is just randomized kinetic energy on a microscopic scale. To gain some insight into this, consider one liter ( \(1 \mathrm{~kg}\) ) of water, and figure out how much energy it would take to heat it from absolute zero temperature \(^{37}\) to \(300 \mathrm{~K}\) assuming that the definition of the calorie (Sec. 5.5; p.73) applies across this entire range. If this same amount of energy went into kinetic energy- hurling the water across the room-what would the corresponding velocity be?

Considering that wind turbines are rated for the maximumtolerable wind speed around \(12 \mathrm{~m} / \mathrm{s}\), and tend to operate at about \(30 \%\) capacity factor, how much average power \(^{44}\) would a \(100 \mathrm{~m}\) diameter turbine operating at \(45 \%\) efficiency be expected to produce?
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