A body has a uniform acceleration of 45km min^ -2 .find its value in ms ^-2 (SI system)
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Answered by
32
accnn= 45km//min ^2
=45km/min.× min.
= 45×1000/60×60
=45000/3600
=450/36
=50/4 = 12.5 m/ s^2
=45km/min.× min.
= 45×1000/60×60
=45000/3600
=450/36
=50/4 = 12.5 m/ s^2
Answered by
32
The dimenstional formula of acceleration is [ L T ^-2 ] .
Let L1 ,T1 denote km and min ,and L2 ,T2 denote m (meter) and s ( second) respectively.
if the numerical values are n1 and n2 respectively, then.
n1 [ L1 T1 ^ -2 ] = n2 [ L2 T2 ^ -2 ]
n2 = n1 [ L1/L2] [T1/T2 ] ^2
hence n1 = 45.
n2 = 45 (km/m) (min_s)^-2
= 45 (1000m/m) (60s/s)^-2
= 45 × 1000/60 × 60
= 12.5
thus the acceleration of the body is 12.5ms^-2
Let L1 ,T1 denote km and min ,and L2 ,T2 denote m (meter) and s ( second) respectively.
if the numerical values are n1 and n2 respectively, then.
n1 [ L1 T1 ^ -2 ] = n2 [ L2 T2 ^ -2 ]
n2 = n1 [ L1/L2] [T1/T2 ] ^2
hence n1 = 45.
n2 = 45 (km/m) (min_s)^-2
= 45 (1000m/m) (60s/s)^-2
= 45 × 1000/60 × 60
= 12.5
thus the acceleration of the body is 12.5ms^-2
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