A, with a charge of−5.0μC, is located 3.0cm to the left of sphereB, with a charge of +3.0μC. a. What is the force of sphereB on sphereA? b. A third sphere, C, with a +2.0 μC charge, is added 5.0cm directly beneathA. What is the new net force on sphereA : Convert μCto C . (x10 -6)Convert cm to m . (x 10-2)

i dont know but

explanation:

whoever your teacher is, deserves a raise for using pokemon as an example

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