Respuesta :
Answer:
[tex] 1.24\times 10^{36}[/tex]
Explanation:
[tex]q[/tex] = magnitude of charge on each proton = 1.6 x 10⁻¹⁹ C
[tex]m[/tex] = mass of each proton = 1.67 x 10⁻²⁷ kg
r = distance between the two protons = 1 x 10⁻¹⁵ m
Electric force between the two protons is given as
[tex]F_{e} = \frac{kq^{2}}{r^{2}}[/tex]
[tex]F_{e} = \frac{(9\times 10^{9})(1.6\times 10^{-19})^{2}}{(1\times 10^{-15})^{2}}[/tex]
[tex]F_{e} = 230.4[/tex] N
Gravitational force between the two protons is given as
[tex]F_{g} = \frac{Gm^{2}}{r^{2}}[/tex]
[tex]F_{g} = \frac{(6.67\times 10^{-11})(1.67\times 10^{-27})^{2}}{(1\times 10^{-15})^{2}}[/tex]
[tex]F_{g} = 1.86\times 10^{-34}[/tex] N
Ratio is given as
[tex]Ratio =\frac{F_{e}}{F_{g}}[/tex]
[tex]Ratio =\frac{230.4}{1.86\times 10^{-34}}[/tex]
[tex]Ratio = 1.24\times 10^{36}[/tex]
Answer:
[tex]F_e:F_g = 1:1.237 * 10^{36}[/tex]
Explanation:
Parameters given:
Distance between the protons, r = [tex]1 * 10^{-15} m[/tex]
Electric charge of proton, Q = [tex]1.602 * 10^{-19} C[/tex]
Mass of proton, m = [tex]1.673 *10^{-27} kg[/tex]
The electric force on one proton due to the other given as:
[tex]F_e = \frac{k*Q*Q}{r^2}[/tex]
where k = Coulomb's constant
[tex]F_e = \frac{9 * 10^9 * 1.602 * 10^{-19} * 1.602 * 10^{-19} }{(1 * 10^{-15})^2} \\\\\\F_e = 230.98 N[/tex]
The gravitational force on one proton due to the other is given as:
[tex]F_g = \frac{G*m*m}{r^2}[/tex]
where G = gravitational constant
[tex]F_g = \frac{6.67408 * 10^{-11}*1.673 *10^{-27} * 1.673 *10^{-27}}{(1 * 10^{-15})^2} \\\\\\F_g = 1.868 * 10^{-34} N[/tex]
Therefore, the ratio of electric force, [tex]F_e[/tex], and gravitational force, [tex]F_g[/tex], is:
[tex]\frac{F_e}{F_g} = \frac{230.98}{1.868 * 10^{-34}} = \frac{1}{1.237 * 10^{36}} \\\\\\F_e:F_g = 1:1.237 * 10^{36}[/tex]
