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

no. bullets are macro objects. wave nature is for light waves, extremely tiny particles. if a light particle was the size of your finger, the bullet would be the size of the universe. so basically, that huge difference in size makes the wave nature of matter insignificant.