if f=ax+bt² where f is force x is distance, t is time then what is dimension of axc/bt²
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here f=ax=bt^2
f=MLT^-1
and x=distance =L
f=ax
MLT^-1=aL
MLT^-1/L=a
MT^-1=a
f=bt^2
MLT^-1=bT^2
MLT^-1/T^2=b
MTMLT^-3=b
now axc/bt^2
c=bt^2/ax
=MLT^-3×T^2/MT^-1×L
MLT^-6/MLT^-1
T^-5
f=MLT^-1
and x=distance =L
f=ax
MLT^-1=aL
MLT^-1/L=a
MT^-1=a
f=bt^2
MLT^-1=bT^2
MLT^-1/T^2=b
MTMLT^-3=b
now axc/bt^2
c=bt^2/ax
=MLT^-3×T^2/MT^-1×L
MLT^-6/MLT^-1
T^-5
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