# Dimensional AnalysisWatch

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Thread starter 4 years ago
#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.

Thank you very much for your help and advice.
0
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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.

Thank you very much for your help and advice.
Re-check your dimensions for acceleration a 0
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Thread starter 4 years ago
#3
(Original post by davros)
Re-check your dimensions for acceleration a Thank you. Would it be LT-2?
0
reply
4 years ago
#4
(Original post by Ggdf)
Thank you. Would it be LT-2?
Correct 0
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Thread starter 4 years ago
#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
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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
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Thread starter 4 years ago
#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
reply
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
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Thread starter 4 years ago
#9
I'm not sure .. Would it just be T?
0
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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
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Thread starter 4 years ago
#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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