Convert m/s to N? Chart Vectors Graphically

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SUMMARY

The discussion centers on graphing vectors given in meters per second (m/s) without the need to convert them into Newtons. The user is provided with three velocity vectors: 15 m/s at 20 degrees, 6 m/s at 90 degrees, and 8 m/s at -135 degrees. It is established that converting m/s to Newtons is unnecessary, as they represent different physical quantities. The negative angle of -135 degrees indicates a clockwise rotation from the positive x-axis.

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Homework Statement


I'm charting vectors graphically and am given: 15m/s at 20 degrees + 6m/s at 90 degrees + 8 m/s at -135degrees. Do I have to convert the m/s into Newtons before graphing? Also, which direction would that -135degrees be in if you start at a point? I have no idea what the negative degrees mean and haven't found help online.

Homework Equations

The Attempt at a Solution


Tried laying it out similar to if I had N and graphed it to scale, but my numbers were way off.
 
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A) No. existing units are just fine.
 
Have you tried drawing a compass and showing the angles?

If you plot x, y or z, positive is one way, and negative is the other way. So if your angles increase from 0 when you rotate counter-clockwise, what happens to the Angle when you go the other way?
 
tvshonk said:

Homework Statement


I'm charting vectors graphically and am given: 15m/s at 20 degrees + 6m/s at 90 degrees + 8 m/s at -135degrees. Do I have to convert the m/s into Newtons before graphing? Also, which direction would that -135degrees be in if you start at a point? I have no idea what the negative degrees mean and haven't found help online.

Homework Equations



The Attempt at a Solution


Tried laying it out similar to if I had N and graphed it to scale, but my numbers were way off.
It makes no sense to convert to Newtons. The to sets of units refer to completely different physical quantities.

Just use a convenient scale factor to lay the velocity vectors out graphically.
 
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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