Show that the variation of atmospheric pressure with altitude is given by P = P 0 e − αy where α = ρ 0 g/P 0 , P 0 is atmospheric pressure at some reference level y = 0, and ρ 0 is the atmospheric density at this level. Assume the decrease in atmospheric pressure over an infinitesimal change in altitude (so that the density is approximately uniform over the infinitesimal change) can be expressed from Equation 14.4 as dP = − pg dy . Also assume the density of air is proportional to the pressure, which, as we will see in Chapter 18, is equivalent to assuming the temperature of the air is the same at all altitudes.
Show that the variation of atmospheric pressure with altitude is given by P = P 0 e − αy where α = ρ 0 g/P 0 , P 0 is atmospheric pressure at some reference level y = 0, and ρ 0 is the atmospheric density at this level. Assume the decrease in atmospheric pressure over an infinitesimal change in altitude (so that the density is approximately uniform over the infinitesimal change) can be expressed from Equation 14.4 as dP = − pg dy . Also assume the density of air is proportional to the pressure, which, as we will see in Chapter 18, is equivalent to assuming the temperature of the air is the same at all altitudes.
Solution Summary: The author explains that atmospheric pressure is a pressure exerted by the weight of the atmosphere.
Show that the variation of atmospheric pressure with altitude is given by P = P0e−αy where α = ρ0g/P0, P0 is atmospheric pressure at some reference level y = 0, and ρ0 is the atmospheric density at this level. Assume the decrease in atmospheric pressure over an infinitesimal change in altitude (so that the density is approximately uniform over the infinitesimal change) can be expressed from Equation 14.4 as dP = −pg dy. Also assume the density of air is proportional to the pressure, which, as we will see in Chapter 18, is equivalent to assuming the temperature of the air is the same at all altitudes.
A skateboarder with his board can be modeled as a particle of mass 80.0 kg, located at his center of mass. As shown in the figure below, the skateboarder starts from rest in a crouching position at one lip of a half-pipe (point). On his descent, the skateboarder moves without friction so
that his center of mass moves through one quarter of a circle of radius 6.20 m.
i
(a) Find his speed at the bottom of the half-pipe (point Ⓡ).
m/s
(b) Immediately after passing point Ⓑ, he stands up and raises his arms, lifting his center of mass and essentially "pumping" energy into the system. Next, the skateboarder glides upward with his center of mass moving in a quarter circle of radius 5.71 m, reaching point D. As he
passes through point ①, the speed of the skateboarder is 5.37 m/s. How much chemical potential energy in the body of the skateboarder was converted to mechanical energy when he stood up at point Ⓑ?
]
(c) How high above point ① does he rise?
m
A 31.0-kg child on a 3.00-m-long swing is released from rest when the ropes of the swing make an angle of 29.0° with the vertical.
(a) Neglecting friction, find the child's speed at the lowest position.
m/s
(b) If the actual speed of the child at the lowest position is 2.40 m/s, what is the mechanical energy lost due to friction?
]
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