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UnkleRhaukus
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@azooo "Two electrons are separated by 1 meter. What is the ratio of the magnitude of the electric force on one electron and the gravitational force on that electron due to the other electron? The charge of an electron is 1.6*10^19C and its mass is 9.1*10^31??
which formuA will use for it?? "
 one year ago
 one year ago
UnkleRhaukus Group Title
@azooo "Two electrons are separated by 1 meter. What is the ratio of the magnitude of the electric force on one electron and the gravitational force on that electron due to the other electron? The charge of an electron is 1.6*10^19C and its mass is 9.1*10^31?? which formuA will use for it?? "
 one year ago
 one year ago

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Mashy Group TitleBest ResponseYou've already chosen the best response.0
which are the formulae you know to calculate the force of gravity between two point masses? and electric force between 2 point charges? :P
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.3
\[\begin{align*}\left\frac{\vec{\mathbf F}_{\text{electric}}}{\vec{\mathbf{F}}_{\text{gravitational}}}\right &=\frac{K\dfrac{q_{e^}q_{e^}}{d^2}}{G\dfrac{m_{e^}m_{e^}}{d^2}} % =\frac KG\left( \frac{q_{p}}{m_{p}}\right)^2 \\ %&=\frac{9.0\times10^9 \left[\frac{\text N\cdot\text m^2}{\text C^2}\right]} %{6.7\times10^{11}\left[\frac{\text{N}\cdot\text m^2}{\text {kg}^2}\right]} %\left(\frac{1.6\times10^{19}[\text C]}{1.7\times10^{27}[\text{kg}]}\right)^2 \\ %&=\left(\frac{9.0\times1.6^2}{6.7\times1.7^2}\right)\frac{10^{92\times19}}{10^{112\times27}} \\ %&=\left(\frac{9.0\times2.56}{6.7\times2.89}\right)\frac{10^{92\times19}}{10^{1154}} \\ %&=1.2\times10^{29+65} \\ %&=1.2\times10^{36} \end{align*} \]
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
there you go!!.. so whats the problem ?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.3
for the rato of the forces between Protons \[\begin{align*} \left\frac{\vec{\mathbf F}_{\text{electric}}}{\vec{\mathbf{F}}_{\text{gravitational}}}\right &=\frac{K\dfrac{q_{p}q_{p}}{d^2}}{G\dfrac{m_{p}m_{p}}{d^2}} =\frac KG\left( \frac{q_{p}}{m_{p}}\right)^2 \\ &=\frac{9.0\times10^9 \left[\frac{\text N\cdot\text m^2}{\text C^2}\right]} {6.7\times10^{11}\left[\frac{\text{N}\cdot\text m^2}{\text {kg}^2}\right]} \left(\frac{1.6\times10^{19}[\text C]}{1.7\times10^{27}[\text{kg}]}\right)^2 \\ %&=\left(\frac{9.0\times1.6^2}{6.7\times1.7^2}\right)\frac{10^{92\times19}}{10^{112\times27}} \\ &=\left(\frac{9.0\times2.56}{6.7\times2.89}\right)\frac{10^{92\times19}}{10^{1154}} \\ %&=1.2\times10^{29+65} \\ &=1.2\times10^{36} \end{align*}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.3
notice how the distance cancels out,
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
oh lol you opened this question to explain it to someone lol.. sorry :D
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.3
can close the question now @azooo ?
 one year ago
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