To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.
For the particular case of the Cube with equal sides its volume is determined by
[tex]V_c = l^3[/tex]
[tex]V_c = 6^3 = 216cm^3[/tex]
In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say
[tex]V_d = \pi r^2*l[/tex]
[tex]V_d = \pi (\frac{2}{2})^2*6[/tex]
[tex]V_d = 6\pi cm^3[/tex]
In this way the net volume would be
[tex]\Delta V = V_c-V_d[/tex]
[tex]\Delta V = 216cm^3-6\pi cm^3[/tex]
[tex]\Delta V = 197.15cm^3 = 197.15*10^{-6}m^3[/tex]
We need to find the mass, but we have the Weight and Gravity so from Newton's second Law
[tex]F= mg[/tex]
[tex]m = \frac{F}{g}[/tex]
[tex]m = \frac{6.6}{9.8}[/tex]
[tex]m = 0.673kg[/tex]
PART A) From the relation of density as a unit of mass and volume we have to
[tex]\rho = \frac{m}{V}[/tex]
[tex]\rho = \frac{0.673}{197.15*10^{-6}}[/tex]
[tex]\rho = 3413.64kg/m^3[/tex]
PART B) To find the weight of the cube then we apply the ratio of
[tex]W = mg[/tex]
[tex]W = V\rho g[/tex]
[tex]W = (216*10^{-6})(3413.64)(9.8)[/tex]
[tex]W = 7.22N[/tex]
a. The density of the metal is ρ = 3420 kg/m³
b. The weight of the cube before you drilled the hole in it is 7.24 N
Since the cube has a length L = 6.0 cm, its volume is V = L³
= (6.0 cm)³
= 216 cm³.
Since the hole is a cylinder and has a diameter of d = 2.0 cm and height h = L = 6.0 cm, its volume V' = πd²L/4
= π(2.0 cm)² × 6.0 cm/4
= π × 4.0 cm² × 6.0 cm/4
= π × 1.0 cm² × 6.0 cm
= 6π cm³
= 18.85 cm³
So, the volume of the cube after drilling the cylindrical hole is V" = V - V'
= 216 cm³ - 18.85 cm³
= 197.15 cm³
The density of the metal is ρ = 3420 kg/m³
The density of the metal ρ = m/V" where
So, ρ = m/V"
ρ = W/gV"
So, substituting the values of the variables into the equation, we have
ρ = W/gV"
ρ =6.60 N/(9.8 m/s² × 197.15 × 10⁻⁶ m³)
ρ = 6.60 N/1932.07 × 10⁻⁶ m⁴/s²)
ρ = 0.00342 × 10⁶ kg/m³
ρ = 3420 kg/m³
So, the density of the metal is ρ = 3420 kg/m³
The weight of the cube before you drilled the hole in it is 7.24 N
W = mg where
Substituting the values of the variables into the equation, we have
W = ρVg
W = 3420 kg/m³ × 216 × 10⁻⁶ m³ × 9.8 m/s²
W = 7239456 × 10⁻⁶ kgm/s²
W = 7.239456 kgm/s²
W = 7.239456 N
W ≅ 7.24 N
So, the weight of the cube before you drilled the hole in it is 7.24 N
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