Question

One cubic meter (1.00 m3) of aluminum has a mass of 2.70 103 kg, and the same volume of iron has a mass of 7.86 103 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 1.82 cm on an equal-arm balance.

Answer 1

Answer:

$r_{al}=2.598cm$

Explanation:

Density =mass/volume

$D_{al}=2.70103kg/m^{3}$

$D_{iron}=7.86103/m^{3}$

condition for balance:

$M_{iron}=M_{al}$

M=D*Volum

then:

$D_{iron}*4/3*pi*r_{ir}^{3}=D_{al}*4/3*pi*r_{al}^{3}$

$r_{al}=r_{ir}*\sqrt[3]{\frac{D_{iron}}{D_{al}}}$

$r_{al}=2.598cm$

Answer 2

Answer:

1.28 cm

Explanation:

Density (ρ) is an intensive property resulting from the quotient of mass (m) and volume (V).

The density of Al is ρAl = 2.70 × 10³ kg/m³

The density of Fe is ρFe = 7.86 × 10³ kg/m³

A Fe sphere of radius 1.82 cm has the following volume.

V = 4/3 × π × r³

V = 4/3 × π × (0.0182 cm)³

V = 2.53 × 10⁻⁵ m³

The mass corresponding to this sphere is:

2.53 × 10⁻⁵ m³ × 2.70 × 10³ kg/m³ = 0.0683 kg

The Al sphere has the same mass, so its volume is:

0.0683 kg × (1 m³/7.86 × 10³ kg) = 8.69 × 10⁻⁶ m³

The radius corresponding to an Al sphere of 8.69 × 10⁻⁶ m³ is:

V = 4/3 × π × r³

8.69 × 10⁻⁶ m³ = 4/3 × π × r³

r = 0.0128 m = 1.28 cm