Answer:
a) 6.4 x 10^-12 cm^3
b) 17 x 10^-6 mm^2
Explanation:
The complete question is...
Bacteria vary somewhat in size, but a diameter of 2.3 m is not unusual.
A. What would be the volume (in cubic centimeters) of such a bacterium, assuming that it is spherical? Express your answer using two significant figures.
B. What would be the surface area (in square millimeters) of such a bacterium, assuming that it is spherical?
Diameter of the bacteria d = 2.3 μm
Radius r = d/2 = 2.3/2 = 1.15 μm
The shape is assumed to be spherical
a) The volume = volume of a sphere = [tex]\frac{4}{3} \pi r^3[/tex]
V = [tex]\frac{4}{3}*3.142* 1.15^3[/tex] = 6.3715 μm^3
1 μm^3 = 10^-12 cm^3
6.3715 μm^3 = 6.3715 x 10^-12 cm^3
==> 6.4 x 10^-12 cm^3
b) The surface area = [tex]4\pi r^2[/tex]
A = 4 x 3.142 x [tex]1.15^2[/tex] = 16.62 μm^2
1 μm^2 = 10^-6 cm^2
16.62 μm^2 = 16.62 x 10^-6 mm^2
==> 17 x 10^-6 mm^2
Answer:
A) V =1.3×10−11cm3
B) S =2.6×10−5mm2
Explanation:
A) The volume V of a sphere in terms of its radius is V=43πr3 . The radius is one-half the diameter or r=d/2=1.45μm . Finally, the necessary equalities for this problem are: 1μm=10−6m ; 1cm=10−2m ; and 1mm=10−3m .
EXECUTE: V=43πr3=43π(1.45μm)3 (10−6m1μm)3 (1cm10−2m)3 =1.3×10−11cm3 .
B) The surface area A of a sphere in terms of its radius is A=4πr2 . The radius is one-half the diameter or r=d/2=1.45μm . Finally, the necessary equalities for this problem are: 1μm=10−6m ; 1cm=10−2m ; and 1mm=10−3m .
EXECUTE: A=4πr2=4π(1.45μm)2 (10−6m1μm)2(1mm10−3m)2 =2.6×10−5mm2
