Question

In order to get going fast, eagles will use a technique called stooping, in which they dive nearly straight down and tuck in their wings to reduce their surface area. While stooping, a 6- kg golden eagle can reach speeds of up to 53 m/s . While golden eagles are not very vocal, they sometimes make a weak, high-pitched sound. Suppose that while traveling at maximum speed, a golden eagle heads directly towards a pigeon while emitting a sound at 1.1 kHz. The emitted sound has a sound intensity level of 30 dB when heard at a distance of 5 m .A) Model this stooping golden eagle as an object moving at terminal velocity. The eagle’s drag coefficient is 0.5 and the density of air is 1.2 kg/m 3 . What is the effective cross-sectional area of the eagle’s body while stooping?B) What is the doppler-shifted frequency that the pigeon will hear coming from the eagle?C) Consider the moment when the pigeon is 5 m away from the eagle. At the pigeon’s position, what is the intensity (in W/m^2 ) of the sound the eagle makes?D) The golden eagle slams into the 250- g pigeon, which is initially moving at 10 m/s in the opposite direction (toward the eagle). The eagle grabs the pigeon in its talons, and they move off together in a perfectly inelastic collision. How fast do they move after the collision?

Answer 1

Check the explanation

Step-by-step explanation:

Part A

F = CA

this drag force balances the weight = 6X 9.8

so

6X9.8 = 0.5 X A X0.5 X 1.2 X 532

A= 0.069 m2

Part B

here the sorce is moving and the observer is at rest

so f= f(- 1 - 1

f = 1.1X10 343 343 – 53

f' = 1.3 KHz

Part C:

given the intensity = 30 dB

we know that I dB = 10 log (I(W/m2))

so we get I (W/m2) = 1000

Part D : The catch

Given that U1 = 53 M1 = 6 kg

U2 =-10 M2=0.25

V1=V2

now conserving momentum

6 X 53 -0.25 X10 =(6+0.25)V

V= 50.48 m/sec