contestada

You have been working on a new, strong, lightweight ceramic which will be used to replace steel bearing balls. One cubic meter of the steel alloy currently in use has a mass of 8.08 ⨯ 10^3 kg, whereas a cubic meter of your new material has a mass of 3.14 ⨯ 10^3 kg. If the balls currently in use have a radius of 1.9 cm and for this application to keep balls of the same mass instead of the same size, what would the radius replacement ball of the new alloy be?

Respuesta :

gcosarinsky

Answer:

r=2.6 cm

Explanation:

Hi!

Lets call steel material 1, and the new alloy material 2. You know their densities:

[tex]\rho_1=8.08*10^3\frac{kq}{m^3}\\\rho_2=3.14*10^3\frac{kq}{m^3}[/tex]

The volume of a sphere with radius r is given by:

[tex]V=\frac{4}{3}\pi r^3[/tex]

Then the masses of the bearings are:

[tex]m_1=\rho_1V_1=\frac{4}{3}\pi r_1^3 \\m_1=\rho_2V_2=\frac{4}{3}\pi r_2^3[/tex]

For the masses to be the same:

[tex]\rho_1 V_1 = \rho_2 V_2\\\frac{V_2}{V_1} =\frac{\rho_1}{\rho_2}\\(\frac{r_2}{r_1})^3 = \frac{8.08}{3.14}=2.57\\r_2=\sqrt[3]{2.57}\;r1 =1.37r_1=1.37*1.9cm=2.6cm[/tex]

Otras preguntas