Conceptual Physics

would be too easy!)

Pa

14.2 A stream of alcohol (density 790 kg/m^3 ) is

flowing through the pipeline shown. The

speeds v 1 and v 2 are 11.7 m/s and 15.8 m/s.

The gauge pressures P 1 and P 2 are

2.77×10^5 Pa and 1.15×10^5 Pa, and the height

h 1 is 3.60 m. What is h 2?

m

14.3 An open can is completely filled with water, to a depth of 20.6 cm. A hole is punched in the can at a height of 1.7 cm from the

bottom of the can. Bernoulli's equation can be used to derive the following formula for the speed of the water flowing from the

hole.

In this formula, h represents the depth of the submerged hole below the surface of the water. (a) How fast does the water

initially flow out of the hole? (b) How fast does the water flow when the can is half empty?

(a) m/s

(b) m/s

Section 16 - The Earth’s atmosphere

16.1 The relationship between pressure and altitude in the Earth’s atmosphere is complicated by the fact that the density of air is

not constant, but decreases with height according to an equation called the law of atmospheres, which we do not show here.

This law assumes that the atmosphere is at a constant overall Kelvin temperature T.

Because of its nonconstant density and its ill-defined upper limit, the pressure of the atmosphere cannot be described in

terms of some atmospheric depth h below the “top” of the atmosphere by a simple equation like P = ȡgh. Instead, the

atmospheric pressure P is a function of the altitude y above the ground, as given by the following equation that can be

derived as an immediate consequence of the law of atmospheres:

In this pressure equation, Patm is atmospheric pressure at sea level (y = 0),m is the average mass in kilograms of an

atmospheric molecule, g is the acceleration due to gravity, and k is Boltzmann’s constant (k = R/NA = 1.38×10–23 J/K).

(a) If the dry atmosphere is composed of 78.1% nitrogen molecules (mN = 28.0 u), 21.0% oxygen molecules

(mO = 32.0 u), and 0.930% argon molecules (mA = 39.9 u), plus trace amounts of other substances, what is the average

mass of an atmospheric molecule in atomic mass units u?

(b) Convert the average atmospheric molecular mass from atomic mass units to kilograms, using the relationship

1 u = 1.66×10–27 kg.

(c) Now use the pressure equation given above, together with an overall atmospheric temperature of T = 264 K (which is

about –9 °C), to find the theoretical pressure of the atmosphere at 1000 m, 5000 m, and 10,000 m (a common cruising

altitude for commercial jetliners).

(a) u

(b) kg

(c) Pa at 1000 m Pa at 5000 m Pa at 10,000 m

(^274) Copyright 2007 Kinetic Books Co. Chapter 13 Problems