Respuesta :
https://brainly.com/question/5027087
Let C, S and H be the diameters of cell, sand and hair
Diameter of a cell: [tex] C=1 \cdot 10^{-5}m[/tex]
Diameter of a grain of sand: [tex]S=2 \cdot10^{-4}m[/tex]
Part A:
[tex]H=0.000025=2.5 \cdot 10^{-5}[/tex] meters, because it takes 5 movements right of the decimal dot, until there is only one digit to its left.
Part B:
we are comparing [tex]S=2 \cdot10^{-4} m [/tex] to [tex] C=1 \cdot10^{-5} m [/tex] ,
[tex] \frac{S}{C}= \frac{2 \cdot 10^{-4}}{1 \cdot 10^{-5}}=\frac{20 \cdot 10^{-5}}{1 \cdot 10^{-5}}=20[/tex]
thus S is 20 times larger than C
Part C
[tex]1m=1 \cdot10^9nm[/tex]
thus,
[tex]H=2.5 \cdot 10^{-5}m=2.5 \cdot 10^{-5} \cdot 1 \cdot 10^9nm=2.5 \cdot 10^{-5+9}=2.5\cdot10^4nm[/tex]
Part D.
[tex]1m=1\cdot 10^9nm\\\\1nm=\displaystyle{ \frac{1m}{10^9}=1\cdot 10^{-9}m} [/tex]
thus 300 nm are written in meters as follows:
[tex]300nm= 300\cdot 1\cdot 10^{-9}m=3\cdot 10^2 \cdot 10^{-9}m=3\cdot 10^{-7}m[/tex]
Let C, S and H be the diameters of cell, sand and hair
Diameter of a cell: [tex] C=1 \cdot 10^{-5}m[/tex]
Diameter of a grain of sand: [tex]S=2 \cdot10^{-4}m[/tex]
Part A:
[tex]H=0.000025=2.5 \cdot 10^{-5}[/tex] meters, because it takes 5 movements right of the decimal dot, until there is only one digit to its left.
Part B:
we are comparing [tex]S=2 \cdot10^{-4} m [/tex] to [tex] C=1 \cdot10^{-5} m [/tex] ,
[tex] \frac{S}{C}= \frac{2 \cdot 10^{-4}}{1 \cdot 10^{-5}}=\frac{20 \cdot 10^{-5}}{1 \cdot 10^{-5}}=20[/tex]
thus S is 20 times larger than C
Part C
[tex]1m=1 \cdot10^9nm[/tex]
thus,
[tex]H=2.5 \cdot 10^{-5}m=2.5 \cdot 10^{-5} \cdot 1 \cdot 10^9nm=2.5 \cdot 10^{-5+9}=2.5\cdot10^4nm[/tex]
Part D.
[tex]1m=1\cdot 10^9nm\\\\1nm=\displaystyle{ \frac{1m}{10^9}=1\cdot 10^{-9}m} [/tex]
thus 300 nm are written in meters as follows:
[tex]300nm= 300\cdot 1\cdot 10^{-9}m=3\cdot 10^2 \cdot 10^{-9}m=3\cdot 10^{-7}m[/tex]