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Scott
Scott, MIT Graduate
Category: Homework
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Experience:  MIT Graduate (Math, Programming, Science, and Music)
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An airplane flies due west at 150km/h with respect to the ...

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An airplane flies due west at 150km/h with respect to the air. There is a wind blowing at 95 km/h to the northeast relative to the ground. What is the plane's speed with respect to the ground?

Welcome back!

You can think of this as a Pythagorean Theorem problem. The hypotenuse is the unknown. One short leg is the 150km/h minus the east component of the wind. The other short leg is the north component of the wind:

east comp: 95*cos(45) -- so the total leg is 150-95*cos(45)

north comp: 95*sin(45)

The magnitude of the velocity will be:

sqrt((150-95*cos(45))^2 + (95*sin(45))^2)

= 106.64 km/h

Let me know if you have any questions,

Scott

Scott and 5 other Homework Specialists are ready to help you
Customer: replied 9 years ago.
Reply to Scott's Post: Can you also answer this one and i will add another $10.
You walk 20m south and 20m eas. Find the magnitude and direction of the resultant displacement both graphically and algebraically. Answer is 28.28 M what is the degrees counterclockwise from east?

Hi there!

I can help you with that as well.

The magnitude is sqrt(20^2 + 20^2) = 28.28 m

The direction is arctan(-20/20) = -45 degrees (45 degrees south of east) (but that is the same as 315 degrees counterclockwise from east).

Graphically:

graphic

Let me know if you have any questions,

Scott

Customer: replied 9 years ago.
Reply to Scott's Post: last one and i already accepted the last two and added in a bonus..
A weather station releases a balloon that rises at a constant 16m/s relative to the air, but there is a wind blowing at 6.0 m/s toward the west. What is the magnitude and direction of the velocity of the balloon? answer 17.07 m/s   bu how many degrees above the horizontal?

Hi again! As far as I can tell, the first answer was accepted and I got the bonus. Thanks!

The magnitude is another Pythagorean Triangle:

sqrt(16^2 + 6^2) = 17.08 m/s

The angle is given by the arctangent:

angle = arctangent(y/x)

arctan(6/16) = 20.56 degrees

Let me know if you have any questions,

Scott

Customer: replied 9 years ago.
That is what i had as well but it came up incorrect. Any other suggestions for that one.. the rest were all right.. does it change with the "above the horizontal"?

Could it be a rounding problem? For example, have you tried it with 20.6 degrees or 20.556 degrees?

-Scott

Customer: replied 9 years ago.
that didn't work either so maybe there is an error in the program.. let me know if you can think of anything else otherwise i will still accept this one because i had the same answer as well and appreciate your help..
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Scott and 5 other Homework Specialists are ready to help you
Customer: replied 9 years ago.
that did not work either.. i am not sure what they are looking for..

Hmmm.... that is strange. Is there anyone you can contact to see if there is a bug in the program?

-Scott

Customer: replied 9 years ago.
i will ask the teacher tom.. thanks for your help.. i will accept the last one as well.. thanks again