Three individual point charges are placed at the following positions in the x-y plane:Q3= 5.0 nC at (x, y) = (0,0);Q2= -3.0 nC at (x, y) = (4 cm, 0); and Q1= ?nC at (x, y) = (2 cm,0);What isthe magnitude, and sign, ofcharge Q1such that the net force exerted on charge Q3, exerted bycharges Q1and Q2, is zero?

Explanation:

net force exerted on charge Q₃, exerted by charges Q₁and Q₂, will be  zero

if net  electric field due to charges Q₁ and Q₂  at origin is zero .

electric field due to Q₂

= 9 X 10⁹ X 3 x10⁹ / .04²

electric field due to Q₁

= 9 X 10⁹ X Q₁ / .02²

For equilibrium

9 X 10⁹ X Q₁ / .02² = 9 X 10⁹ X 3 x10⁻⁹ / .04²

Q₁  = 3 X10⁻⁹ x .02² / .04²

= 3 / 4 x 10⁻⁹

.75 x 10⁻⁹  C

i already answered it by 34/s

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft