Physics, 02.03.2020 21:03 almighty196

Determine the ratio of the electrostatic force to the gravitational force between a proton and an electron, FE/FG. Note: k = 8.99 × 109 N-m2/C2; G = 6.672 × 10–11 N-m2/kg2; me = 9.109 ×

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Answer from: nicki76

$\frac{F_e}{F_g} = 2.3 \times 10^{18}$

Explanation:

The gravitational force is given by Newton's Law of Gravity:

$F_g = \frac{Gm_1m_2}{r^2}$

The electrostatic force is given by Coulomb's Law:

$F_e = \frac{kq_1q_2}{r^2}$

The ratio between these two forces is

$\frac{F_e}{F_g} = \frac{\frac{kq_1q_2}{r^2}}{\frac{Gm_1m_2}{r^2}} = \frac{kq_1q_2}{Gm_1m_2} = \frac{(8.99\times 10^{-12})(1.6\times 10^{-19})(1.6 \times 10^{-19})}{(6.67\times 10^{-11})(9.1 \times 10^{-31})(1.6 \times 10^{-27})} = 2.3 \times 10^{18}$

Answer from: LittleNerd2852

Complete Question:

Determine the ratio of the electrostatic force to the gravitational force between a proton and electron, FE/FG.

Note : k = 8.99 × 10⁹ N.m²/C² ; G = 6.672 x 10⁻¹¹ Nm²/kg²; me = 9.109 × 10⁻³¹ kg and mp = 1.672 × 10⁻²⁷kg.

FE/FG = 2.3 x 10³⁹

Explanation:

According to Coulomb's law, the electrostatic force ( $F_{E}$ ) between two particles is given as;

$F_{E}$ = k x $\frac{Q_1 * Q_2}{r^2 }$ --------------------(i)

Where;

k = electric constant = 8.99 x 10⁹Nm²/C²

Q₁ = the charge of particle 1

Q₂ = the charge of particle 2

r = the distance of separation between the two particles

Also, according to Newton's law of gravitational force, the gravitational force ( $F_{G}$ ) between two particles is given as;

$F_{G}$ = G x $\frac{M_{1} * M_2}{r^{2} }$ --------------------(ii)

Where;

G = gravitational constant = 6.672 x 10⁻¹¹Nm²/kg²

M₁ = mass of particle 1

M₂ = mass of particle 2

r = distance of separation between the two particles

For clarity, we will calculate $F_{E}$ and $F_{G}$ separately before finding their ratio.

From the question;

The particles are a proton (particle 1) and an electron (particle 2) with the following details;

Q₁ = charge of proton = 1.6 x 10⁻¹⁹C

Q₂ = charge of electron = -1.6 x 10⁻¹⁹C

M₁ = mass of proton = mp = 1.672 × 10⁻²⁷kg

M₂ = mass of electron = me = 9.109 × 10⁻³¹kg

Substitute the values of k, Q₁ and Q₂ into equation (i) as follows;

$F_{E}$ = 8.99 x 10⁹ x $\frac{(1.6*10^{-19}) * ( -1.6* 10^{-19})}{r^2 }$

$F_{E}$ = 8.99 x 10⁹ x $\frac{(2.56*10^{-38})}{r^2 }$ [negative sign can be discarded]

$F_{E}$ = $\frac{(23.01*10^{-29})}{r^2 }$

Also, substitute the values of G, M₁ and M₂ into equation (ii) as follows;

$F_{G}$ = 6.67 x 10⁻¹¹ x $\frac{(1.672*10^{-27}) * (9.109* 10^{-31})}{r^2 }$

$F_{G}$ = 6.67 x 10⁻¹¹ x $\frac{(15.23*10^{-58})}{r^2 }$

$F_{G}$ = $\frac{(101.58*10^{-69})}{r^2 }$

The ratio $F_{E}$ / $F_{G}$ is therefore;

$F_{E}$ / $F_{G}$ = $\frac{(23.01*10^{-29})}{r^2 }$ / $\frac{(101.58*10^{-69})}{r^2 }$

$F_{E}$ / $F_{G}$ = $\frac{(23.01*10^{-29})}{(101.58*10^{-69})}$

$F_{E}$ / $F_{G}$ = ${(0.23*10^{40})}$

$F_{E}$ / $F_{G}$ = ${(2.3*10^{39})}$

Therefore, the ratio is 2.3 x 10³⁹

Answer from: Quest

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is

answer

1,500

kg • m/s.

Answer from: Quest

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