A spherical balloon is inflating with helium at a rate of 192pi ft^3/min. How fast is the balloon's radius increasing at the instant the radius is ft?

3ft/min

Explanation:

Complete question:

A spherical balloon is inflating with helium at a rate of 192pi ft^3/min. How fast is the balloon's radius increasing at the instant the radius is 4ft?

Volume of a sphere = where:

r is the radius of the sphere:

Using the chain rule to find dV/dt we have:

is the rate at which the balloon's radius is increasing

is the rate at which the volume is increasing

Given

radius = 4ft

Required

From the formula:

Get

Hence the balloon radius is increasing at the rate of 3ft/min.

protons in

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards