Dimensional Analysis of an object

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The discussion focuses on determining the dimensions of coefficients A and B in the equation v = At^3 - Bt, where v represents velocity. It is established that velocity has dimensions of length per time, [L/T]. To find the dimensions of A and B, the relationship A*t^3 = [L/T] leads to A being [L/T^4] and B being [L/T^2]. The conversation emphasizes the importance of manipulating units to achieve the desired dimensional form. Overall, understanding dimensional analysis is crucial for solving such physics problems effectively.
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Homework Statement


The speed, v, of an object is given by the equation v=At^3 - Bt, where t refers to time. What are the dimensions of A and B?

Homework Equations


None, but I know that to find dimensions, we look at the units that make up the variable, like the area of a room is [L^2]. I understand the basic concept, but have trouble applying it.

The Attempt at a Solution


v = At^3 - Bt.
I would think that since velocity is a measure of distance per time, we should get t to divide the distance. Could it be then that A is [L/T^4] and B= [L/T^2]?
I can only figure out this way to get T as the denominator in a way that could get A and B added up to get [L/T] for the answer...

Thanks for your help!
 
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Yes, this looks right because you know velocity should have dimensions of displacement/unit time. So you need a displacement in each of the coefficients, and also units of time that gives an inverse proportion in the coefficients.

So, even though this can be done mentally, for other problems you would start with your original proportion and work towards making into the dimension you want; like so:

A*t^3 = \frac{L}{t}

Now divide both sides by t^3.

A = \frac{L}{t^4}

Or if you wanted to convert from km/h to m/s, you know that 1000m=1km and 60s=1m and 60m=1h, so:

\frac{km}{h} * \frac{1h}{60m} * \frac{1m}{60s} = \frac{1}{60^2} \frac{km}{s}

So you see that the hours cancel and make minutes, then the minutes cancel and make seconds in the denominator. Now take care of the km units:

\frac{km}{h} * \frac{1h}{60m} * \frac{1m}{60s} * \frac{1000m}{1km} = \frac{1000}{60^2} \frac{m}{s}

Just write out your proportions and see that the units cancel to get what you want.
 
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Thanks a lot!
 
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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