Answer
(a) The density of this metal is $5900~kg/m^3$
(b) The cube weighed 7.21 N before the hole was drilled.
Work Step by Step
(a) We can find the volume of the cube.
$V_c = (0.050~m)^3 = 1.25\times 10^{-4}~m^3$
We can find the volume of the hole.
$V_h = \pi~r^2~h$
$V_h = (\pi)(0.010~m)^2(0.050~m)$
$V_h = 1.57\times 10^{-5}~m^3$
We can find the volume of the material after the hole is drilled.
$V = V_c - V_h$
$V = (1.25\times 10^{-4}~m^3)-(1.57\times 10^{-5}~m^3)$
$V = 1.09\times 10^{-4}~m^3$
We can find the mass of the material after the hole is drilled.
$M = \frac{weight}{g}$
$M = \frac{6.30~N}{9.80~m/s^2}$
$M = 0.643~kg$
We can find the density of this metal.
$\rho = \frac{M}{V}$
$\rho = \frac{0.643~kg}{1.09\times 10^{-4}~m^3}$
$\rho = 5900~kg/m^3$
The density of this metal is $5900~kg/m^3$
(b) We can find the mass of the material that was removed by the hole.
$M_h = \rho~V$
$M_h = (5900~kg/m^3)(1.57\times 10^{-5}~m^3)$
$M_h = 0.0926~kg$
We can find the weight of this mass.
$weight = (0.0926~kg)(9.80~m/s^2)$
$weight = 0.907~N$
We can find the original weight of the cube.
$weight = 6.30~N + 0.907~N = 7.21~N$
The cube weighed 7.21 N before the hole was drilled.
