The Skin of a Can

Q. A closed can is a cylinder of radius 2.0 cm and height 10.0 cm. What is its whole surface area, in square metres? expects 0.015

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

Inputs

radius

r

2 cm

height

h

10 cm

Outputs

area

S_{c}

The solution

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 Ruby

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

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

1require_relative "../solutions/skin_of_a_can"23play_with :skin_of_a_can do4  title "The Skin of a Can"5end

The laws

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