A current of 50.0A
50.0A flows through a cross-section of a wire. After six hours, what is the total mass of all of the electrons which have passed through the cross-section? Give your answer in milligrams (mg) and to three significant figures.
Solution.
I=50.0A;I=50.0A;I=50.0A;
t=21600s;t=21600s;t=21600s;
q=It;q=It;q=It;
q=50.0⋅21600=1080000C;q=50.0\sdot21600=1080000C;q=50.0⋅21600=1080000C;
q=Ne ⟹ N=qe;q=Ne\implies N=\dfrac{q}{e};q=Ne⟹N=eq;
N=10800001.6⋅10−19=675000⋅1019;N=\dfrac{1080000}{1.6\sdot10^{-19}}=675000\sdot10^{19};N=1.6⋅10−191080000=675000⋅1019;
m=Nme;m=Nm_e;m=Nme;
m=675000⋅1019⋅9.1⋅10−31=6.143⋅10−6kg=6.143mg;m=675000\sdot10^{19}\sdot9.1\sdot10^{-31}=6.143\sdot10^{-6}kg=6.143mg;m=675000⋅1019⋅9.1⋅10−31=6.143⋅10−6kg=6.143mg;
Answer: m=6.143mg.m=6.143 mg.m=6.143mg.
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