Help with dimensional analysis question?

Question

Consider the equation v=1/3zxt^2. The dimensions of the variables x, v, and t are [L], [L]/[T], and [T], respectively. What must be the dimensions of the variable z, such that both sides of the equation have the same dimensions?

Can someone work through the problem so I can understand the process?

Here's where I get stuck:

[L]/[T]=1/3z[L][T]^2
then the L's cancel out, and one of the T's on each side. So you're left with:

nothing=1/3z[T]

so now what?

Or am I even doing that part right?

Answer 1

Harlow

z = 3v/(xt^2)
therefore

z = {[L]/[T]}/([L][T^2])

therefore
z must have units
L/LT^3
1/T^3

NOTE YOU SAY ONE OF THE T'S ON EACH SIDE CANCELS OUT
THat IS INCORRECT sorry for the caps

Answer 2

this sounds like your homework, so I'm not going to answer it for you, but I can walk you through how to approach it. :)

The idea behind dimensional analysis is that you're not just multiplying numbers, you're multiplying physical quantities (like a speed, a mass, a length, etc.) These quantities have a dimension, sometimes called units.

- A length quantity is in units of length (feet, meters, etc.,) which you can think of as [L].
- A time quantity is in units of time (seconds, minutes, etc.), which you can think of as [T].
- A speed quantity is in units of length per unit time (feet/second, miles/hour, etc.), which you can think of as [L]/[T]

When you multiply or divide two quantities, you multiply and divide both their values AND their units. So 1 foot / 2 seconds = 1/2 foot/sec. = 0.5 [L]/[T]

When you write out an equation, you're saying that the stuff on the left side is equal to the stuff on the right. Since we're dealing with quantities, both sides have to have the same units (dimensions). So if all the units of the stuff on the right-hand side of an equation multiply out to units of [T] / ([L]^2) then the units of all the stuff on the left-hand side has to multiply out to units of [T] / ([L]^2).

For example:

if

v/z= (1/5)a/b^2
where v is in [L]/[T],
z is in [T], and
a is in [L]

then on the left you have:

([L]/[T]) / [T] = [L] / [T]^2

and on the right you have:
[L] / b^2


so b^2 must have the units [T]^2

so b must have the units [T]


Does that help? Sorry if I've confused you more, but try to get this. I've caught a LOT of mistakes on tests by checking the units.

Answer 3

Did you mean - v=zxtt/3 (lose the multiply signs, exponentiation, etc.)

In dimensions: (z= ?) (3 has no dimension)
L/T = ?LTT

So we reduce - to state the obvious, divide both sides by TT
L/TTT = ?L
And divide both sides by L
1/TTT = ? so z is T^(-3)

If you meant v=1/(3zxtt) then dimensionally:
L/T = 1/?LTT
Multiply both sides by:
1/T and get L=1/?LT
? and get ?L = 1/LT
1/L and get ? = 1/LLT (Answer #2)

It's been a few years since I did dimensional analysis.

Answer 4

v = 1/3zxt^2
isolate z.

z = 3v / (xt^2)

substitute

z = 3(L/T) / (LT^2)

z = 3 / (L^2T)

rearranging to read easier

z = 3/(TL^2)

Answer 5

I'll start you off.

By substitution:
L/T = (1/3)zLT^2

Answer 6

I think a little T and A would improve your dimention to the degree that you wouldn't need to analize it!

Answer 7

E=MC squared

Answer 8

I don't do math.