When you put water in a kitchen blender, it begins to travel in a 6-cm-radius circle at a speed of 2 m/s. How hard must the sides of the blender push inward on 0.004 kg of the spinning water?

To find the force exerted by the sides of the blender on the spinning water, we can start by calculating the acceleration experienced by the water. We'll use the formula for centripetal acceleration:

a = v^2 / r

where:
a = centripetal acceleration
v = velocity of the water
r = radius of the circle

Let's substitute the given values into the formula:

a = (2 m/s)^2 / 0.06 m

a = 4 m^2/s^2 / 0.06 m

a ≈ 66.67 m/s^2

Now we can calculate the force exerted on the water using Newton's second law of motion:

F = m * a

where:
F = force
m = mass of water
a = acceleration

Substituting the given values:

F = 0.004 kg * 66.67 m/s^2

F ≈ 0.267 kg m/s^2

Therefore, the sides of the blender must push inward with a force of approximately 0.267 Newtons (N) on the spinning water.