1. hey guys.... im new to this forum as well as to the Science field.. Please help me with this physics sum.....

Given that F=a t^(-1) + b r^2 ,where F denotes force & t, time the dimensions of a & b are respectively,

i) L T^(-2) & T^(-2)
ii) T & T^(-2)
iii) L T^(-1) & T^(-2)
iv) ML T^(-3) & ML T^(-4)
v) L^3 T^(-3) & M L^2 T^(-2)

Please tell me how to find the values of 2 unknown terms (a & b) in one equation.......
Thanks !

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3. Read your question again. You don't need to give the values, you need to give the dimensions only. That makes it much easier.

4. Originally Posted by KALSTER
Read your question again. You don't need to give the values, you need to give the dimensions only. That makes it much easier.
oh yeah sorry typing mistake! i wanna knw how to find dimensions of 2 terms in 1 equation... if it was only F = at^(-1) , i could make 'a' the subject & find the answer.. bt there are 2 terms.... please tell me what to do.....

5. This is a question about dimensional analysis. You can consider each term in the sum separately. The dimensions of force are in something like Newtons, you need the dimensions in each of the summands to match that of the dimensions of force(I'm trying not to give this part away). Consider the first term, the units of a multiplied by the units of 1/t must be the same as the units of force. I'm assume M means a unit of mass, T a unit of time and L a unit of length. The same can be done for the second term. I hope this helps!

6. Originally Posted by TheObserver
This is a question about dimensional analysis. You can consider each term in the sum separately. The dimensions of force are in something like Newtons, you need the dimensions in each of the summands to match that of the dimensions of force(I'm trying not to give this part away). Consider the first term, the units of a multiplied by the units of 1/t must be the same as the units of force. I'm assume M means a unit of mass, T a unit of time and L a unit of length. The same can be done for the second term. I hope this helps!
yeah got it... Thanks a lot!!

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