The volume of a balloon an be approximated by V = 4 3 π r 3 V=43πr3. If air is leaking from the balloon at a rate of 56 cubic centimeters per second, how fast is the radius of the balloon shrinking at the moment the radius is 5 centimeters? Make sure to choose the correct units for your answer.

The rate  the radius of the balloon shrinking at the moment the radius is 5 centimeters is 0.1783 cm/s

Explanation:

Here we have

dV/dt = 56 cm³/s

When the radius is 5 cm we have

Therefore,

From which,

The rate  the radius of the balloon shrinking at the moment the radius is 5 centimeters = 0.1783 cm/s.

Radius is increasing at the rate of 0.1783 cm/s

Explanation:

Let r be the radius of the balloon and V be its volume at any time t.

Thus, V= (4/3)πr³

Now, let's differentiate both sides with respect to t and we obtain;

dV/dt = 4πr²•(dr/dt)

From the question, we are given that; dV/dt = 56 cm³/s

Thus,

56 = 4πr²•(dr/dt)

dr/dt = 56/(4πr²)

So,we want to find dr/dt at r=5,

Thus,

dr/dt = 56/(4π•5²)

dr/dt = 0.56/π = 0.1783 cm/s

answer: (431,2)

step-by-step explanation:

an energy transformation flow diagram is shown

where is it ?