An electrically charged point particle with a mass of 1.41 x 10-6 kg is launched horizontally into a uniform electric field with a magnitude of 686 N/C. The particle's initial velocity is 5.39 m/s, directed eastward, and the external electric field is also directed eastward. A short time after being launched, the particle is 9.89 x 10-3 m below and 0.151 m east of its initial position. Determine the net charge carried by the particle, including the algebraic sign ( or -).

the net charge carried by the particle is -2.35 × 10⁻⁷ C

Explanation:

Given that;

mass m = 1.41 × 10⁻⁶ kg

Magnitude E = 686 N/C

v = 5.93 m/s

A short time after being launched, the particle is 9.89 x 10-3 m below and 0.151 m east of its initial position

consider vertical;

u =0, y = 9.89 × 10⁻³ m, g = a = 9.8 m/s²

Now, from the second equation of motion;

s = ut + 1/2 × at²

so, y = ut + 1/2 × at²

we substitute

9.89 × 10⁻³ = 0×t + (1/2 × 9.8 × t²)

9.89 × 10⁻³ = 4.9t²

t² = 9.89 × 10⁻³ / 4.9

t² = 0.0201836

t = √0.00201836

t = 0.0449 s

Consider horizontal;

u = 5.93 m/s, a = (qE/m) m/s, t = 0.0449 s, x = 0.151 m

also, from the second equation of motion;

s = ut + 1/2 × at²

so

x =  ut + 1/2 ×  (qE/m) × t²

we substitute

0.151 =  (5.93 × 0.0449) + (1/2 ×  ((q×686)/1.41 × 10⁻⁶) × (0.0449 )²

0.151 = 0.266257 +  0.69149143q / 1.41 × 10⁻⁶

0.151 - 0.266257 =  0.69149143q / 1.41 × 10⁻⁶

-0.115257 = 0.69149143q / 1.41 × 10⁻⁶

-0.115257 × 1.41 × 10⁻⁶ = 0.69149143q

0.69149143q = -1.6251237 × 10⁻⁷

q = -1.6251237 × 10⁻⁷ / 0.69149143

q = -2.35 × 10⁻⁷ C

Therefore, the net charge carried by the particle is -2.35 × 10⁻⁷ C

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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i would think it's food chain