You are told that in a thundercloud, a 0.85 g raindrop can acquire a charge of 1.1 mc. a) assume that two such raindrops are separated by 3.9 cm. what is the acceleration, in meters per square second, of each raindrop away from each other? b) the numbers provided seem to be reasonable - why should you not believe the source of this information? a. the acceleration is far too small. b. the raindrops would not stay at a constant distance from each other, so this situation would never occur. c. the acceleration is far too large. d. raindrops are not this size. c) which premise or assumption is responsible for the error? a. the raindrop is too large. b. none of these. c. the mass is too large for a raindrop. d. the assumed charge of the raindrops are too small. e. the assumed charges of the raindrops are too large. f. the mass is too small for a raindrop.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

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