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$1J=1kg.m^2/s^2$. To convert this to $erg=g.cm^2/s^2$ replace $kg$ by $10^3 g$ and $m$ by $10^2 cm$ : $$1J=1(10^3 g)(10^2 cm)^2/s^2 = 10^7 g.cm^2/s^2=10^7 erg$$


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As Pangloss points out, this is surprisingly non-trivial! The sort of manipulations you are talking about are part of a quantity calculus. This is not to be confused with Newton's and Leibniz's Calculus. A calculus is just a method for computation. This is a method of manipulating quantities which yield "the same" quantity with different units. ...


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The question seems trivial, but it isn't at all! Some strange formal properties of algebraic calculations on units (e.g. the elusive rad unit which appears or disappears) should suggest that "something is wrong" instead of the more usual W.Allen's "Whatever works". The problem concerns the logic (syntax and semantics) of the ''...


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Good question. I don't have any elegant, deep mathematical answer for you. But let's consider an object moving at 3 m/sec and consider a time interval of 5 sec. We can lay out the object's progress along a line consisting of 5 sections, in each of which the object has moved 3 m. So the total distance moved is 3m + 3m +3m +3m +3m = 15m. Knowing multiplication,...


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