Dimensional Analysis Watch

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

Name:  dimentional analysis.jpg
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Size:  47.7 KB

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:

Name:  Acceleration.jpg
Views: 100
Size:  63.0 KB


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.
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davros
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(Original post by Ggdf)
Please can anyone offer me feedback on whether I've answered these dimensional analysis questions (see attached picture) correctly?

Name:  dimentional analysis.jpg
Views: 98
Size:  47.7 KB

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:

Name:  Acceleration.jpg
Views: 100
Size:  63.0 KB


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
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Ggdf
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(Original post by davros)
Re-check your dimensions for acceleration a
Thank you. Would it be LT-2?
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davros
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(Original post by Ggdf)
Thank you. Would it be LT-2?
Correct
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Ggdf
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(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?
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davros
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(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!!
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Ggdf
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(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
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davros
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(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 T^{-2} \times T^2 ?
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Ggdf
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(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?
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davros
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(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
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Ggdf
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(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
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!
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