# Dimensional AnalysisWatch

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#1
Please can anyone offer me feedback on whether I've answered these dimensional analysis questions (see attached picture) correctly?

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] [s]^-1

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

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

?

I hope this makes sense.

0
4 years ago
#2
(Original post by Ggdf)
Please can anyone offer me feedback on whether I've answered these dimensional analysis questions (see attached picture) correctly?

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] [s]^-1

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

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

?

I hope this makes sense.

Re-check your dimensions for acceleration a
0
#3
(Original post by davros)
Re-check your dimensions for acceleration a
Thank you. Would it be LT-2?
0
4 years ago
#4
(Original post by Ggdf)
Thank you. Would it be LT-2?
Correct
0
#5
(Original post by davros)
Correct
Thank you so much .

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?
0
4 years ago
#6
(Original post by Ggdf)
Thank you so much .

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!!
0
#7
(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
0
4 years ago
#8
(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
And what do you get if you work out ?
0
#9
(Original post by davros)
And what do you get if you work out ?
I'm not sure .. Would it just be T?
0
4 years ago
#10
(Original post by Ggdf)
I'm not sure .. Would it just be T?
No, this is just the normal law of indices applied to dimensions.

So in this case you have i.e. there is no T dependence! So all you are left with in your equation is L, which is what you want
0
#11
(Original post by davros)
No, this is just the normal law of indices applied to dimensions.

So in this case you have i.e. there is no T dependence! So all you are left with in your equation is L, which is what you want
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!
0
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