At a particular sea level location at a particular time the atmospheric pressure is 14.8 lbs/in^2 and the temperature is 80 degrees F. Estimate the temperature, pressure, and air density at an altitude of 8,000 ft.

30.52 °F

1572 lbs/ft²

0.0019 slug/ft³

Explanation:

Given that

Pressure at sea level, P₁ = 14.8 PSI

Temperature, T₁ = 80 °F

At 8,000 ft (m) altitude

for altitude h, less than 36000 ft, the model used to modify temperature and pressure is written below

T = 59 - 0.00356h, where h is the altitude

T = 59 - 0.00356(8,000)

T = 59 - 28.48

T = 30.52 °F

Next, we find the pressure which is calculated using the formula

P = 2116 [(T + 459.7)/518.6]^5.256

P = 2116 [(30.52 + 459.7) / 518.6]^5.256

P = 2116 [490.22/518.6]^5.256

P = 2116 (0.945)^5.256

P = 2116 * 0.743

P = 1572 lbs/ft²

The density is gotten using the formula

ρ = P / [1718 (T + 459.7)]

ρ = 1572 / [1718 (30.52 + 459.7)]

ρ = 1572 / (1718 * 490.22)

ρ = 1572 / 842198

ρ = 0.0019 slug/ft³

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop