Question

Assuming Biological substances are 90 percent water and the density of water is 1.0 x 10^3 kg/m^3, estimate teh masses (density multiplied by volume) of the following.
a. a spherical cell with a diameter of 1.0 um (volume = 4/3 pi(r^3)
b. a fly, which can be approximated by a cylinder 4.0 mm long and 2.0 mm in diameter (volume = 1pi(r^2)

Answer 1

A) The answer is 5.2 · 10⁻¹⁴ kg

It is known that the mass (m) is density (D) multiplied by volume (V):
m = D · V
D = 1.0 · 10³ kg/m³
m = ?
V = ?

Let's calculate the volume.
The volume (V) of a spherical cell is V = 4/3 π r³        (r - radius of a sphere)
r = ?     
d = 1.0 um = 1.0 · 10⁻⁶ m
Since radius is a half of the diameter, then:
r = d ÷ 2 = 1.0 · 10⁻⁶ m ÷ 2 = 0.5 · 10⁻⁶ m
It is known than π = 3.14
Therefore:
V = 4/3 π r³ = 4/3 · 3.14 · (0.5 · 10⁻⁶)³ = 4/3 · 3.14 · 0.5³ · (10⁻⁶)³
                    = 4/3 · 3.14 · 0.125 · 10⁻⁶*³ = 4/3 · 3.14 · 0.125 · 10⁻¹⁸ = 0.52 · 10⁻¹⁸
                    = 5.2 · 10⁻¹⁷ m₃

So, now when we know D and V, it is easy to calculate m:
m = D · V = 1.0 · 10³ kg/m³ · 5.2 · 10⁻¹⁷ m³ = 5.2 · 10⁻¹⁷⁺³ kg = 5.2 · 10⁻¹⁴ kg


b) The answer is 12.56 · 10⁻⁶ kg
It is known that the mass (m) is density (D) multiplied by volume (V):
m = D · V
D = 1.0 · 10³ kg/m³
m = ?
V = ?

Let's calculate the volume.
The volume (V) of a fly in the shape of cylinder is V = h π r²
h = 4.0 mm = 4.0 · 10⁻³ m
r = ?     
d = 2.0 mm = 2.0 · 10⁻³ m
Since radius is a half of the diameter, then:
r = d ÷ 2 = 2.0 · 10⁻³ m ÷ 2 = 1.0 · 10⁻³ m
It is known than π = 3.14
Therefore:
V = h π r³ = 4.0 · 10⁻³ · 3.14 · (1.0 · 10⁻³ )² = 4.0 · 10⁻³ · 3.14 · 1.0² · (10⁻³ )² =
                 = 4.0 · 10⁻³ · 3.14 · 1.0 · 10⁻³*² = 4.0 · 10⁻³ · 3.14 · 1.0 · 10⁻⁶ =
                 = 12.56 · 10⁻³⁻⁶ = 12.56 · 10⁻⁹ m³

So, now when we know D and V, it is easy to calculate m:
m = D · V = 1.0 · 10³ kg/m³ · 12.56 · 10⁻⁹ m³ = 12.56 · 10⁻⁹⁺³ kg = 12.56 · 10⁻⁶ kg