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)

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