rate of change = At an altitude of feet above sea level, the air pressure, , in

Question

rate of change =

At an altitude of feet above sea level, the air pressure, , in inches of mercury (in Hg), is given by An unpressurized seaplane takes off at an angle of to the horizontal and a speed of 115 mph. What is the rate of change of pressure in the plane with respect to time at take-off, in inches of mercury per second? (If your answer contains trig functions not evaluated to a decimal value, use degrees.)

rate of change =

At an altitude of h feet above sea level, the air pressure, P, in inches of mercury (in Hg), is given by An unpressurized seaplane takes off at an angle of 45^{circ} to the horizontal and a speed of 115 mph. What is the rate of change of pressure in the plane with respect to time at take - off, in inches of mercury per second? (If your answer contains trig functions not evaluated to a decimal value, use degrees.) rate of change =

Explanation / Answer

vertical Component of take off velocity = 115 * cos 45 mph = 81.3172 mph

Now P = 30 e^(-3.23 * 10^-5 * h)

or h = ( ln(P) - ln (30) ) / (-3.23 * 10^-5)

Differentiate wrt time

or dh/dt = 1/(-3.23 * 10^-5) * (1/P) * dP/dt

Now dh/dt = rate of change of height wrt time = 81.3172 mph

So 81.3172 = 1/(-3.23 * 10^-5) * (1/P) * dP/dt

or dP/dt = -0.002626 * P

Now P at take of = 30 * e^ 0 because height is 0

So P = 30 inches of Hg

dP/dt = -0.002626 * 30 = -0.07879 inches of Hg

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