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26 Nov 2019
Why does step 3 inquestion 44 in ch 14 in the book Collegephysics (8th) By Youngcompute (.5kg)(.39*10^3 J/kg.K)=.195??????
taking .39*10^3 out of scientific notation makes it 390.
so (390)(.5) =195 !!!! right??? I dont get it. someone explainit tome.will rate lifesaver
Why does step 3 inquestion 44 in ch 14 in the book Collegephysics (8th) By Youngcompute (.5kg)(.39*10^3 J/kg.K)=.195??????
taking .39*10^3 out of scientific notation makes it 390.
so (390)(.5) =195 !!!! right??? I dont get it. someone explainit tome.
taking .39*10^3 out of scientific notation makes it 390.
so (390)(.5) =195 !!!! right??? I dont get it. someone explainit tome.
will rate lifesaver
Patrina SchowalterLv2
11 Aug 2019
