Year 12s: what will be the first way you research your future options? >>

Dimensional Analysis

Please can anyone offer me feedback on whether I've answered these dimensional analysis questions (see attached picture) correctly?

dimentional analysis.jpg

Dimensions =
[length] = L
[time] = T
[mass] = M
[number] = 1

i) d = 1/2 at^2

d = L
a = L^2T^-2
t = T

1/2 at ^2 = [1/2][a][t]
= 1 L^2T^-2
= L^2T^-2?

Not well formed??

NB: I assumed that [a] was L^2T^-2 after looking at this:

ii) Is the answer mv? Because ...

[M] = L
[v] = [ms^-1]
[v] = [m] [s^-1]
[v] = [m] :undefined:

^-1

[mv] = [m][v]
= MLT^-1

Also:
[v] = [distance]/[time]
=L/T = LT^-1

?

I hope this makes sense.

Thank you very much for your help and advice.

Reply 1

davros

[QUOTE="Ggdf;55070303"]Please can anyone offer me feedback on whether I've answered these dimensional analysis questions (see attached picture) correctly?

dimentional analysis.jpg

ii) Is the answer mv? Because ...

[M] = L
[v] = [ms^-1]
[v] = [m] [s^-1]
[v] = [m] :^-1:

Re-check your dimensions for acceleration a :smile:

Reply 2

Ggdf

Original post by davros

Re-check your dimensions for acceleration a :smile:

Thank you. Would it be LT-2?

Reply 3

davros

Original post by Ggdf

Thank you. Would it be LT-2?

Correct

Reply 4

Ggdf

Original post by davros

Correct

Thank you so much :smile:

.

Would this be correct?

i) d = 1/2 at^2

d = L
a = LT-2
t = T

1/2 at ^2 = [1/2][a][t]
= 1 LT-2
= LT-2?

Would it still not be well formed as the equation is not the same on both sides?

Reply 5

davros

Original post by Ggdf

Thank you so much :smile:

.

Would this be correct?

i) d = 1/2 at^2

d = L
a = LT-2
t = T

1/2 at ^2 = [1/2][a][t]
= 1 LT-2
= LT-2?

Would it still not be well formed as the equation is not the same on both sides?

You don't seem to have done anything with the dimensions of t^2!!

Reply 6

Ggdf

Original post by davros

You don't seem to have done anything with the dimensions of t^2!!

d = 1/2 at^2

d = L
a = LT-2
t = T

1/2 at ^2 = [1/2][a][t]
= 1 LT-2 T^2
= LT-2T^2

?

Thank you :smile:

Reply 7

davros

Original post by Ggdf

d = 1/2 at^2

d = L
a = LT-2
t = T

1/2 at ^2 = [1/2][a][t]
= 1 LT-2 T^2
= LT-2T^2

?

Thank you :smile:

And what do you get if you work out

T^{-2} \times T^2

Reply 8

Ggdf

Original post by davros

And what do you get if you work out

T^{-2} \times T^2

I'm not sure .. Would it just be T?

Reply 9

davros

Original post by Ggdf

I'm not sure .. Would it just be T?

No, this is just the normal law of indices

a^m \times a^n = a^{m+n}

applied to dimensions.

So in this case you have

T^2 \times T^{-2} = T^0 = 1

i.e. there is no T dependence! So all you are left with in your equation is L, which is what you want :smile:

Reply 10

Ggdf

Original post by davros

No, this is just the normal law of indices

a^m \times a^n = a^{m+n}

applied to dimensions.

So in this case you have

T^2 \times T^{-2} = T^0 = 1

i.e. there is no T dependence! So all you are left with in your equation is L, which is what you want :smile:

Oh, I see what you mean! Therefore it is well formed as d [L] = [L]?

You have been incredibly helpful and kind to assist me with this. I'm really grateful! :smile: