Write dimensions of ab in the relation E = (b-x2) /at, where E is the energy, x is the distance and t is the time.
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Answered by
23
we can only subtract those who have same dimension so...thats how u start
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[ML^2T^-2]= [L]/a[T]. .......... a= [L][T^-1]/[ML^2T^-2]........=[M^-1L ^ -1 T] ...for b= [L]
im sorry i didnt knew i was just posting cropped answers
Answered by
67
we are given that E=b-x2/at
due to principal of homogeneity b=E i.e [M L2 T-2]
similarly x2/at= E i.e [M L2 T-2]
so at= x2/[M L2 T-2]
at=[L2]/[M L2 T-2]
at=[M-1L0T2]
a=[M-1 T2]/[T]
a= [M-1 T1]
so a*b =[M-1T1]*[M L2 T-2]
a*b= [L2 T-1]
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