A Hair Made Large

solved Dimensions

Q. Under a microscope that makes lengths 100 $m_{l} = 100$ times larger, a hair looks 3.5 mm $l_{i} = 3.5$ mm wide. How wide is the hair, in metres?

Inputs

magnification

m_{l}

100

seen width

l_{i}

3.5 mm

Outputs

width

l_{o}

How it's solved

Solved with Dimensions

Given

$m_{l} = 100$
$l_{i} = 3.5$ mm

Find $l_{o}$ .

Working

$l_{i} = 3.5$ mm $= 0.0035$ m
lengths magnified $m_{l} = \frac{l_{i}}{l_{o}}$
$100 = \frac{0.0035}{l_{o}}$
$l_{o} = 3.5 \times 10^{−5}$ m
$l_{o} \approx 3.5 \times 10^{−5}$ mto 2 significant figures

Conclusion

A magnification multiplies lengths; it does not care what the lengths are of.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_hair_made_large do4  kind :numerical5  describes "Under a microscope that makes lengths %{magnification} times larger, a hair looks %{seen_width} wide. %{width}, in metres?"6  given magnification: 100.07  given seen_width: "3.5 mm"8  asks width: "How wide is the hair"9end

solution.rb

1require_relative "../questions/a_hair_made_large"2require_relative "../science/units_and_measurement/dimensions"34solve :a_hair_made_large, using: Dimensions do5  take magnification: :m_l, seen_width: :l_i6  take width: :l_o7  answer find_value(:width, figures: 2)89  conclusion "A magnification multiplies lengths; it does not care what the lengths are of."10end

playground.rb

1require_relative "../solutions/a_hair_made_large"23play_with :a_hair_made_large do4  title "A Hair Made Large"5end

The laws it drew on

quantity	written	units	within	from
speed of light	$c$	m/s	—	Units
avogadro constant	$N_{A}$	mol⁻¹	—	Units
length	$l$	m	0 … ∞	Units
mass	$m$	kg	—	Units
time	$t$	s	0 … ∞	Units
electric current	$I$	A	—	Units
temperature	$T$	K	—	Units
amount of substance	$n$	mol	—	Units
luminous intensity	$I_{v}$	cd	—	Units
arc length	$s$	m	—	Units
radius	$r$	m	0 … ∞	Units
plane angle	$θ$	—	—	Units
area	$A$	m²	—	Units
solid angle	$Ω$	—	—	Units
light time	$t_{light}$	s	—	Units
light distance	$d_{light}$	m	—	Units
number of entities	$N$	—	—	Units
baseline	$B$	m	—	Units
distance away	$D$	m	—	Units
parallax angle	$θ_{p}$	—	—	Units
pi	$π$	—	0 … ∞	Dimensions
gravitational constant	$G$	N m² kg⁻²	—	Dimensions
volume	$V$	m³	—	Dimensions
speed	$v$	m/s	—	Dimensions
acceleration	$a$	m/s²	—	Dimensions
force	$F$	N	—	Dimensions
density	$ρ$	kg/m³	—	Dimensions
work	$W$	J	—	Dimensions
surface area	$S$	m²	—	Dimensions
height	$h$	m	—	Dimensions
breadth	$b$	m	—	Dimensions
cylinder surface area	$S_{c}$	m²	—	Dimensions
volume of a slab	$V_{s}$	m³	—	Dimensions
relative density	$ρ_{rel}$	—	—	Dimensions
density of water	$ρ_{w}$	kg/m³	—	Dimensions
linear magnification	$m_{l}$	—	—	Dimensions
image length	$l_{i}$	m	—	Dimensions
object length	$l_{o}$	m	—	Dimensions
image area	$A_{i}$	m²	—	Dimensions
object area	$A_{o}$	m²	—	Dimensions
volume of them all	$V_{N}$	m³	—	Dimensions
molar volume	$V_{m}$	m³	—	Dimensions
room to spare	$R_{V}$	—	—	Dimensions

θ = \frac{s}{r}

arc over radius

Ω = \frac{A}{r^{2}}

area over radius squared

d_{light} = c \cdot t_{light}

distance at light speed

N = n \cdot N_{A}

entities in moles

θ_{p} = \frac{B}{D}

baseline over distance

V = l^{3}

volume of a cube

v = \frac{l}{t}

average speed

a = \frac{v}{t}

average acceleration

F = m \cdot a

newtons second law

ρ = \frac{m}{V}

mass per volume

W = F \cdot l

force through a distance

S = 6 \cdot l^{2}

surface of a cube

V = \frac{4 \cdot π \cdot r^{3}}{3}

volume of a sphere

S_{c} = 2 \cdot π \cdot r \cdot (r + h)

surface of a cylinder

A = l \cdot b

area of a rectangle

V_{s} = l \cdot b \cdot h

volume from three sides

ρ_{rel} = \frac{ρ}{ρ_{w}}

density against water

m_{l} = \frac{l_{i}}{l_{o}}

lengths magnified

{m_{l}}^{2} = \frac{A_{i}}{A_{o}}

areas magnified

V_{N} = N \cdot V

atoms side by side

R_{V} = \frac{V_{m}}{V_{N}}

molar over atomic volume