Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 10250 by jaikar last updated on 31/Jan/17

Commented by jaikar last updated on 31/Jan/17

$$pleaseQno.10$$$$$$

Answered by ridwan balatif last updated on 31/Jan/17

$$10.\mathrm{n}=1.5625\times {10}^{18}\mathrm{electron},1\mathrm{electron}\mathrm{has}\approx 1.6\times {10}^{-19}\mathrm{C}$$$$\mathrm{Q}/\mathrm{t}=\mathrm{n}.\mathrm{e}/1\mathrm{s}$$$$=1.5625\times {10}^{18}\times 1.6\times {10}^{-19}$$$$=2.5\times {10}^{-1}\mathrm{C}/\mathrm{s}$$$$\mathrm{t}=\frac{12\mathrm{C}}{2.5\times {10}^{-1}\mathrm{C}/\mathrm{s}}=48\mathrm{s}$$$$\mathrm{Answer}:\mathrm{C}$$

Commented by jaikar last updated on 31/Jan/17

$$thanks$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com