# Dimensional Analysis and constant

• ment4ll
In summary, Homework Equations state that V-A=B/t+Ct^4, V-A-B/t=Ct^4 and (V-A-B/t)/t^4=C. When adding quantities, the quantities must be equal in order for them to be added. C has to be a measureable quantity in order for volume to exist. When dimensions are written in terms of standard units (L,T,A,K,M), the result is [L3], [L3t], and [L3/t^4].

## Homework Statement

The volume of an object is given as a function of time by
V= A + B/t + Ct^4

Find the dimension of the constant C.

N/A

## The Attempt at a Solution

I don't understand dimensional analysis completely .. I understand it on simpler terms, but in terms of this problem I do not.

V - A = B/t + Ct^4
V - A - B/t = Ct^4
(V - A - B/t)/t^4 = C

is that correct?

Dimension analysis is just to find its units essentially.

When you are adding, the quantities must be the same. So if you have volume of on the left, A,B/t and Ct^4 all must have combined units of volume.

If that is the case, then what is the units of C?

also note that 2 quantities can only be added when their dimentions are equal.

what can you conclude from this?

rock.freak667 said:
Dimension analysis is just to find its units essentially.

When you are adding, the quantities must be the same. So if you have volume of on the left, A,B/t and Ct^4 all must have combined units of volume.

If that is the case, then what is the units of C?

I mean, yea I get that A, B/t and Ct^4 all are combined units of volume, they have to be. I don't understand how you get the units of C if it isn't the answer I posted.

Sorry if I'm noob at this, I've never taken a physics class in my life (besides this one).

cupid.callin said:
also note that 2 quantities can only be added when their dimentions are equal.

what can you conclude from this?

So C has to be something measurable to volume right?

as A can be added to B, both have same dimensions... let's say [X]

now [X]/[T] = [C][T]4 = [V] = [M]3

can you find dim of C as well as A,B now?

cupid.callin said:
as A can be added to B, both have same dimensions... let's say [X]

now [X]/[T] = [C][T]4 = [V] = [M]3

can you find dim of C as well as A,B now?

Sorry guy, I'm just not grasping it at this point .. wouldn't the dimensions of A+B = X then? I'm confused about where the M3 comes from ..

dimensions is just like witing units in form of 7 basic standard units
what is the unit of 5m/s + 3m/s ??

M is the dimensional symbol for mass

7 dimentions:

Length: [L]

Mass: [M]

Time: [T]

Electric current: [A]

Temperature: [K]

Luminous intensity: [Cd]

Amount of substance: [Mol]

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cupid.callin said:
dimensions is just like witing units in form of 7 basic standard units
what is the unit of 5m/s + 3m/s ??

M is the dimensional symbol for mass

7 dimentions:

Length: [L]

Mass: [M]

Time: [T]

Electric current: [A]

Temperature: [K]

Luminous intensity: [Cd]

Amount of substance: [Mol]

Okay, the unit is m/s and the dimensions is L/T, now I understand that.

GOOD!

now that you are familiar with dimensions and that they don't change while addition or subtraction, can you now write the dimensions of A,B,C ??

Well I am assuming that C is Length while A is electric current and B is mass .. I'm trying to comprehend it man, but that is just guessing based on what you've explained :\

ment4ll said:
Well I am assuming that C is Length while A is electric current and B is mass\

Well how do you got these?

cupid.callin said:
Well how do you got these?

wait, would the deminsions of A, B, and C all be L3?

ment4ll said:
wait, would the dimensions of A, B, and C all be L3?
No, but the dimensions of A, B/t, and C·t4 are all [L]3, because each has to have the same dimension, which is the same dimension as V, which is also [L]3.

I told toy that treat dimentions just like units.

Ok what is unit of A in: (A-B)/t = 1kg ?

SammyS said:
No, but the dimensions of A, B/t, and C·t4 are all [L]3, because each has to have the same dimension, which is the same dimension as V, which is also [L]3.

Thats what I didn't know

now I see that A=L^3, B=L^3t and C=L^3/t^4

I didn't know that you had to cancel out to get the dimension and I didn't know that V=L^3

V=L^3 because the formula for volume=lwh and those are 3 lengths multiplied together.. is that right?

Yes you are right but still A is not L3

Remember what you did ... You know Ct4 + <something> = volume

Now using table you find out that dimension of volume is [L3]

You also know that any quantity X can only be added to some quantity which has same units as X ...
(thats why 5m/s + 3m/s = 8 m/s ... but if someone says that find 5m/s + 3m/s2 ... you tell them to go home and study!)

... By all this can you conclude that Ct4 has same dimentions as that of volume i.e. [L3] ...?

So Dimension of C comes out to be [L3]/[T4]

Now can you find the correct dimensions of A ?