The Skin of a Can

solved Dimensions

Q. A closed can is a cylinder of radius 2.0 cm $r = 2.0$ cm and height 10.0 cm $h = 10.0$ cm. What is its whole surface area, in square metres?

Inputs

radius

r

2 cm

height

h

10 cm

Outputs

area

S_{c}

How it's solved

Solved with Dimensions

Given

$r = 2.0$ cm
$h = 10.0$ cm

Find $S_{c}$ .

Working

$r = 2.0$ cm $= 0.02$ m
$h = 10.0$ cm $= 0.1$ m
surface of a cylinder $S_{c} = 2 \cdot π \cdot r \cdot (r + h)$
$S_{c} = 2 \cdot 3.14159265 \cdot 0.02 \cdot (0.02 + 0.1)$
$S_{c} = 0.0150796447$ m²
$S_{c} \approx 0.015$ m²to 2 significant figures

Conclusion

The radius, with fewest figures, decides how many the area can have.

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 :skin_of_a_can do4  kind :numerical5  describes "A closed can is a cylinder of radius %{radius} and height %{height}. %{area}, in square metres?"6  given radius: "2.0 cm"7  given height: "10.0 cm"8  asks area: "What is its whole surface area"9end

solution.rb

1require_relative "../questions/skin_of_a_can"2require_relative "../science/units_and_measurement/dimensions"34solve :skin_of_a_can, using: Dimensions do5  take radius: :r, height: :h6  take area: :S_c7  answer find_value(:area, figures: 2)89  conclusion "The radius, with fewest figures, decides how many the area can have."10end

playground.rb

1require_relative "../solutions/skin_of_a_can"23play_with :skin_of_a_can do4  title "The Skin of a Can"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