Step-by-step explanation:
I assume the measuring cylinder measures the cm³ of its normally liquid content.
because for anything else we would need more information about the dimensions of the cylinder.
and so, originally the cylinder shows 40 cm³.
after dropping the stone in, how many cm³ of water is the cylinder then showing ?
let's first mention some facts we are going to use :
water weighs 1 kg (1000 g) per liter.
and 1 liter fits exactly into a cube of
10 cm × 10 cm × 10 cm = 1000 cm³
so, 1 cm³ water weighs exactly 1 g and has therefore a density of 1 g / cm³.
the stone has a density of 8.6 g / cm³, is therefore heavier than water and sinks (and replaces water correspondingly).
how many cm³ does the stone have (and replaces water) ?
well, it has 129 g, and 8.6 g of the stone fill a cm³.
so, it has
129 / 8.6 = 15 cm³
therefore, as these 15 cm³ of stone replace 15 cm³ of water, this is the same as putting 40 + 15 = 55 cm³ of water into the measuring cylinder.
and the cylinder reads now 55 cm³.
FYI : but there is still only 40 cm³ of water in there.
this is actually used to calculate the density of objects (by first weighing and then dropping them into the water to see how much water they replace).
Answer: 55 cm³.
Step-by-step explanation:
[tex]\displaystyle\\m_{stone}=129\ g\ \ \ \ \ \rho_{stone}=8,6\ \frac{g}{cm^3} \ \ \ \ \ V_{water}=40\ cm^3\ \ \ \ \ V=?.\\ \boxed {\rho=\frac{m}{V} }\\V=\frac{m}{\rho} \\V=V_{water}+V_{stone}\\V_{stone}=\frac{m_{stone}}{\rho_{stone}} \\V_{stone}=\frac{129}{8,6}\\V_{stone }=15\ cm^3.\\V=40+15\\V=55\ cm^3.[/tex]
