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Question Number 13831 by tawa tawa last updated on 24/May/17

Answered by ajfour last updated on 24/May/17

$$\left(5\right).$$$$F_e=\frac{e^2}{4\pi {ϵ}_0{r}^2};F_g=G\frac{m_e{m}_p}{r^2}$$$$\frac{F_e}{F_g}=\left(\frac{e^2}{m_e{m}_p}\right)\left(\frac{1}{4\pi {ϵ}_{0}G}\right)$$$$=\frac{{(1.6\times {10}^{-19})}^2\left(9\times {10}^9\right)}{(9.1\times {10}^{-31})(1.7\times {10}^{-27})(6.67\times {10}^{-11})}$$$$=\frac{2.56\times 9}{9.1\times 1.7\times 6.67}\times {10}^{-38+9+31+27+11}$$$$\approx \frac{\left(5/2\right)}{\left(5/3\right)\left(20/3\right)}\times {10}^{40}$$$$\approx \frac{9}{40}\times {10}^{40}or=2.2\times {10}^{39}.$$$$\left(6\right).$$$$\alpha particlehaschargeq=+2e$$$$F=\frac{\left(2e\right)\left(e\right)}{4\pi {ϵ}_0{r}^2}$$$$F=\frac{2{(1.6\times {10}^{-19})}^2\left(9\times {10}^9\right)}{{\left({10}^{-13}\right)}^2}$$$$=2\times 2.56\times 9\times {10}^{-38+9+26}$$$$=5.12\times 9\times {10}^{-3}N$$$$F=4.608\times {10}^{-2}N\approx 0.46mN.$$

Commented by tawa tawa last updated on 24/May/17

$$\mathrm{God}\mathrm{will}\mathrm{always}\mathrm{help}\mathrm{you}\mathrm{too}.\mathrm{Thank}\mathrm{you}\mathrm{sir}.$$

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