GO Adiabatic wind . The normal airflow over the Rocky Mountains is west to east. The air loses much of its moisture content and is chilled as it climbs the western side of the mountains, When it descends on the eastern side, the increase in pressure toward lower altitudes causes the temperature to increase. The flow, then called a chinook wind, can rapidly raise the air temperature at the base of the mountains. Assume that the air pressure p depends on altitude y according to p = p 0 exp (− ay), where p 0 = 1.00 atm and a = 1.16 × 10 −4 m −1 . Also assume that the ratio of the molar specific heats is γ = 4 3 A parcel of air with an initial temperature of −5.00°C descends adiabatically from y 1 = 4267 m to y = 1567 m. What is its temperature at the end of the descent?
GO Adiabatic wind . The normal airflow over the Rocky Mountains is west to east. The air loses much of its moisture content and is chilled as it climbs the western side of the mountains, When it descends on the eastern side, the increase in pressure toward lower altitudes causes the temperature to increase. The flow, then called a chinook wind, can rapidly raise the air temperature at the base of the mountains. Assume that the air pressure p depends on altitude y according to p = p 0 exp (− ay), where p 0 = 1.00 atm and a = 1.16 × 10 −4 m −1 . Also assume that the ratio of the molar specific heats is γ = 4 3 A parcel of air with an initial temperature of −5.00°C descends adiabatically from y 1 = 4267 m to y = 1567 m. What is its temperature at the end of the descent?
GO Adiabaticwind. The normal airflow over the Rocky Mountains is west to east. The air loses much of its moisture content and is chilled as it climbs the western side of the mountains, When it descends on the eastern side, the increase in pressure toward lower altitudes causes the temperature to increase. The flow, then called a chinook wind, can rapidly raise the air temperature at the base of the mountains. Assume that the air pressure p depends on altitude y according to p = p0 exp (−ay), where p0 = 1.00 atm and a = 1.16 × 10−4 m−1. Also assume that the ratio of the molar specific heats is
γ
=
4
3
A parcel of air with an initial temperature of −5.00°C descends adiabatically from y1= 4267 m to y = 1567 m. What is its temperature at the end of the descent?
Find the ratio of the diameter of silver to iron wire, if they have the same resistance per unit length (as they might in household wiring).
d
Ag
= 2.51
dFe
×
Show that the units 1 v2/Q = 1 W, as implied by the equation P = V²/R.
Starting with the equation P = V²/R, we can get an expression for a watt in terms of voltage and resistance. The units for voltage, V, are equivalent to [?
v2
v2
A, are equivalent to J/C ✓ X . Therefore, 1
= 1
= 1 A V1 J/s
Ω
V-A X
= 1 W.
. The units for resistance, Q, are equivalent to ?
The units for current,
Please solve and answer the question correctly please. Thank you!!
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