91影视

• The side length of the cube is,$a=5.0\mathrm{cm}\left(\frac{1m}{100\mathrm{cm}}\right)=0.05\mathrm{m}$
• The diameter of the drill is, $D=2.0\mathrm{cm}\left(\frac{1m}{100\mathrm{cm}}\right)=0.02\mathrm{m}$
• The weight of the cube after the hole is drilled was, W = 6.3 N

The density of a substance is defined as the amount of mass contained inside a unit volume.

The density can be given as,

$蚁=\frac{m}{V}$鈥︹赌︹赌︹赌︹赌︹赌�..(1)

Where m and V are the mass and volume of the substance.

The mass of the metal can be evaluated by,

$\begin{array}{r}m=\left(\frac{W}{g}\right)\\ =\frac{6.3\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2}\\ =0.64\mathrm{N}/\left(\mathrm{m}/{\mathrm{s}}^2\right)\\ =0.64\mathrm{kg}\end{array}$

The volume of the cube after the drill can be evaluated as,

$\begin{array}{r}V=V_{\text{cube}}鈭�V_{\text{hole}}\\ =(0.05\mathrm{m}{)}^3鈭�蟺(0.01\mathrm{m}{)}^2\left(0.05\mathrm{m}\right)\\ =1.09脳{10}^{鈭�4}{\mathrm{m}}^3\end{array}$

Now, the density of metal can be evaluated using equation (1),

$\begin{array}{r}蚁=\frac{0.64\mathrm{kg}}{1.09脳{10}^{鈭�4}{\mathrm{m}}^3}\\ =5.9脳{10}^3\mathrm{kg}/{\mathrm{m}}^3\end{array}$

Thus, the density of the metal is $5.9脳{10}^3\mathrm{kg}/{\mathrm{m}}^3$.

The mass of the cube can be evaluated by rearranging equation (1),

$m=蚁V_{cube}$......................(2)

Substitute the values in equation (2),

$\begin{array}{r}m=\left(5.9脳{10}^3\mathrm{kg}/{\mathrm{m}}^3\right)(0.05\mathrm{m}{)}^3\\ m=0.74\mathrm{kg}\end{array}$

The weight of the drill hole can be evaluated by,

$\begin{array}{r}W=mg\\ =\left(0.74\mathrm{kg}\right)脳\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\\ =7.25\mathrm{kg}鈰�\mathrm{m}/{\mathrm{s}}^2\left(\frac{1\mathrm{N}}{1\mathrm{kg}鈰�\mathrm{m}/{\mathrm{s}}^2}\right)\\ =7.25\mathrm{N}\end{array}$

Thus, the weight of the cube is 7.25N