Problem with Coriolis Acceleration

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The discussion focuses on calculating the Coriolis acceleration for a plane flying at 600 mph at a specific latitude and elevation. The relevant equation for Coriolis acceleration is aCor = 2Ω × vxy, where Ω represents the Earth's rotation rate and vxy is the velocity of the plane. The Earth's rotation rate is given as 0.0000727 rad/s, and the plane's velocity is converted to 268.22 m/s. There is an emphasis on using proper vector notation and ensuring calculations are presented clearly, especially regarding significant figures. The conversation highlights the importance of showing work in homework problems for effective assistance.
dennis wang
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Homework Statement


Find the magnitude in g's of the Coriolis acceleration due to the Earth's rotation of a plane flying 600 mph due north over the location of latitude (37˚13'04"N), longitude (121˚50'39"W), at an elevation of 80 meters above mean sea level.

Homework Equations


aCor = 2Ω × vxy

The Attempt at a Solution


n/a
 
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On homework problems, you will have to show some work to get help. What values or calculations can you give for ##\Omega## and ##v_{xy}##?
 
FactChecker said:
On homework problems, you will have to show some work to get help. What values or calculations can you give for ##\Omega## and ##v_{xy}##?

##\Omega## = Earth's rotation=0.0000727
##v_{xy}## = 268.22 m/s
 
dennis wang said:
##\Omega## = Earth's rotation=0.0000727
##v_{xy}## = 268.22 m/s
Sorry, these are vectors. I should have used a notation that made that clear. The better representation is
##\vec a_{Cor} = 2\vec \Omega \times \vec v##
(I'm not sure right now what your xy subscript is for)
 
dennis wang said:
##\Omega## = Earth's rotation=0.0000727
When using three significant figures, the relevant rotation rate in radians per second would relate to the sidereal day, not to the solar day.

[Though it is not clear that the "600 mph" parameter is good to three significant figures]
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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