### Video Transcript

A bowling ball has a mass of 5.5 kilograms. The bowling ball is a sphere with a radius of seven centimeters. What is the density of the bowling ball? Give your answer to the nearest kilogram per cubic meter.

In this question, we’re told that we’ve got a spherical bowling ball and that this ball has a radius of seven centimeters. Let’s label this radius as 𝑟. Now, there are a couple of things we should briefly make clear about the shape of the bowling ball. We’ve drawn the ball here, represented by a circle, but of course, in reality, it’s a sphere which is a three-dimensional shape. When we see a bowling ball in the real world, they typically have three holes on them so that they can be picked up. However, in this question, we’re simply told that the ball is a sphere, so we don’t need to worry about these holes. The other bit of information that we’re given is that the bowling ball has a mass, which we’ll label as 𝑚, that’s equal to 5.5 kilograms. We’re being asked to work out the ball’s density.

So, let’s recall that the density of an object, which is typically labeled as 𝜌, is equal to the object’s mass 𝑚 divided by the object’s volume 𝑉. For our bowling ball, we already know the value of its mass, but at the moment we don’t know its volume. However, we do know that the ball is a sphere. And we can recall that the volume of a sphere is equal to four-thirds times 𝜋 times 𝑟 cubed, where 𝑟 is the sphere’s radius. Since we know the radius of our sphere, we can go ahead and substitute it into this equation to calculate the sphere’s volume.

However, before we do this, let’s notice that we’re asked to give our density in units of kilograms per cubic meter. That means that we’re going to want a value for the volume 𝑉 in units of cubic meters. Since our radius is currently in units of centimeters, then we’re going to want to convert it into meters before we substitute it into this equation. We can recall that one meter is equal to 100 centimeters. Or if we divide both sides by 100, then on the right-hand side, 100 divided by 100 is simply one. And so, we have that one centimeter is equal to one-hundredth of a meter. So, to convert this radius 𝑟 from units of centimeters into units of meters, we need to divide the value by 100.

In units of meters then, we have that 𝑟 is equal to seven divided by 100 meters. This works out at 0.07 meters. We can now take this value for 𝑟 and substitute it into this equation to calculate the sphere’s volume. We get that the volume 𝑉 is equal to four-thirds times 𝜋 times the cube of 0.07 meters. The cube of 0.07 meters works out as 0.000343 cubic meters. Then, multiplying this value by 𝜋 and four over three, we calculate a volume of 0.001437 cubic meters, where the ellipses here are used to indicate that the result has further decimal places. So, we now know the value of both the mass 𝑚 of the bowling ball and its volume 𝑉.

Since the mass is in units of kilograms and the volume is in cubic meters, then when we calculate the density using these values, we will indeed get a result in units of kilograms per cubic meter. So, let’s take our values for 𝑚 and 𝑉 and substitute them into this equation for the density 𝜌. Then, we have that the density is equal to the value of 5.5 kilograms that we were given for the ball’s mass 𝑚 divided by this value that we’ve calculated for its volume 𝑉.

Evaluating the expression gives a result of 3828.071 etcetera kilograms per cubic meter. Since we’re told to give our answer to the nearest kilogram per cubic meter, then we need to take this result in kilograms per cubic meter and round it to the nearest whole number. When we do this, it gives us our final answer to the question that to the nearest kilogram per cubic meter, the density of the bowling ball is equal to 3828 kilograms per cubic meter.