check the correctness of the equations equal to x= Vot+1/2 at square
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Explanation:
Check the correctness of s = UT +1/2at2
s is distance so it's dimensions become L.
Nd other side we have ut+1\2at^2.
as 1\2 is a constant it will have dimensions and apply the dimensions to other quantities.
on solving u will get L there also i,e ur LHS = RHS.
thus the equation is dimensionally consistent.
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Answered by
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Explanation:
Your question:-
- Check the correctness of the given equation
Your solution:-
Given equation- x= Vot+1/2at^2
Dimensions of each terms:-
x=[M^0 L^1 T ^0]
Vo=[M^0 L^1 T^-1]
a=[M^0 L^1 T^-2]
T=[M^0 L^0 T^1]
Implementing given dimension in the equation:-
x= Vot+1/2at^2
[L]=[LT^-1]×[T]+1/2[LT^-2] [T^2]
→ {Anything to the power (1)=given value)} [Note]
After multiplying all dimensions we got:-
[L]=[L]+[L]
Hence LHS=RHS. So the given equation is dimensionally correct.
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