Volume/Mass/Density Problem

[wallace13]

An irregularly shaped chunk of concrete has a hollow spherical cavity inside. The mass of the chunk is 34 kg, and the volume enclosed by the outside surface of the chunk is 0.035 m3. What is the radius of the spherical cavity?


D=m/v
4/3 pi r^3= volume of a sphere


2200 kg/ m cubed (density of concrete)= 34 kg/ v
V= .01545
.035-.01545= .01955
.01955= 4/3pi r^3
r=.167 m

 

[LowlyPion]

Looks OK.

 

[thomate1]

Based on the given information, it appears that the radius of the spherical cavity is approximately 0.167 meters. This calculation was determined by using the formula for density (D=m/v) and the formula for the volume of a sphere (V=4/3 pi r^3). By substituting the given values for mass and volume, the volume of the sphere was calculated to be 0.01545 m3. Subtracting this from the total volume of the chunk (0.035 m3) gave the volume of the hollow spherical cavity (0.01955 m3). By rearranging the formula for the volume of a sphere and substituting the calculated value for volume, the radius was determined to be 0.167 meters. However, it is important to note that this is an approximation as the chunk is described as "irregularly shaped" and may not have a perfectly spherical cavity. Further measurements or calculations may be needed to determine the exact radius.