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Convert `1 MW` power on a new system having basic units of mass , length , and time as `10 kg , 1 dm , and 1 min` , respectively . |
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Answer» `[P] = [ML^(2)T^(-3)]` Using the relation , `n_(2) = n_(1) [(M_(1))/(M_(2))]^(x) [(L_(1))/(L_(2))]^(y) [(T_(1))/(T_(2))]^(z)` ` = 1xx 10^(6) [( 1kg)/( 10 kg) ]^(1) [ ( 1 m)/( 1 dm)]^(2) [ ( 1 s)/( 1 min) ]^(-3)` `[ As 1 MW = 10^(6) W ]` ` = 10^(6) [( 1 kg)/( 10 kg)] [ ( 10 dm)/( 1dm)]^(2) [ ( 1 s)/( 60 s)]^(-3) = 2.16 xx 10^(12) unit` |
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