play phryzby with physics

A Calorie Elsewhere

solved Dimensions

Q. A calorie is 4.2 JW=4.2J. In a system whose units of mass, length and time are α kg, β m and γ s, how many of the new units of energy is a calorie?

How it's solved

Solved with Dimensions

Given

  • W=4.2J

Find W.

Working
  1. W is [M L² T⁻²]as Dimensions says
  2. its new unit is (α kg)(β m)²(γ s)⁻² = α β² γ⁻² J
  3. so W = 4.2 J = 4.2 α⁻¹ β⁻² γ² new units
Conclusion

A unit changes as its dimensions say. Energy carries mass once, length twice and time to the minus two, so the new unit of energy is that mixture of the new base units, and a calorie is its old number divided by it.

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_calorie_elsewhere do4  kind :scq5  describes "A calorie is %{calorie}. In a system whose units of mass, length and time are α kg, β m and γ " \6          "s, how many of the new units of energy is a calorie?"7  given calorie: "4.2 J"89  option "4.2 α⁻¹ β⁻² γ²", as: :inverse10  option "4.2 α β² γ⁻²", as: :direct11  option "4.2 α⁻¹ β⁻¹ γ", as: :half_inverse12  option "4.2 α β γ", as: :plain13end
solution.rb
1require_relative "../questions/a_calorie_elsewhere"2require_relative "../science/units_and_measurement/dimensions"34solve :a_calorie_elsewhere, using: Dimensions do5  take calorie: :W6  take inverse: "4.2 α-1 β-2 γ2", direct: "4.2 α β2 γ-2", half_inverse: "4.2 α-1 β-1 γ", plain: "4.2 α β γ"7  answer find_value(:W, in: { M: "α kg", L: "β m", T: "γ s" })89  conclusion "A unit changes as its dimensions say. Energy carries mass once, length twice and " \10             "time to the minus two, so the new unit of energy is that mixture of the new base " \11             "units, and a calorie is its old number divided by it."12end
playground.rb
1require_relative "../solutions/a_calorie_elsewhere"23play_with :a_calorie_elsewhere do4  title "A Calorie Elsewhere"5end
The laws it drew on
quantitywrittenunitswithinfrom
speed of lightcm/s—Units
avogadro constantNAmol⁻¹—Units
lengthlm0 … ∞Units
massmkg—Units
timets0 … ∞Units
electric currentIA—Units
temperatureTK—Units
amount of substancenmol—Units
luminous intensityIvcd—Units
arc lengthsm—Units
radiusrm0 … ∞Units
plane angleθ——Units
areaAm²—Units
solid angleΩ——Units
light timetlights—Units
light distancedlightm—Units
number of entitiesN——Units
baselineBm—Units
distance awayDm—Units
parallax angleθp——Units
piπ—0 … ∞Dimensions
gravitational constantGN m² kg⁻²—Dimensions
volumeVm³—Dimensions
speedvm/s—Dimensions
accelerationam/s²—Dimensions
forceFN—Dimensions
densityρkg/m³—Dimensions
workWJ—Dimensions
surface areaSm²—Dimensions
heighthm—Dimensions
breadthbm—Dimensions
cylinder surface areaScm²—Dimensions
volume of a slabVsm³—Dimensions
relative densityρrel——Dimensions
density of waterρwkg/m³—Dimensions
linear magnificationml——Dimensions
image lengthlim—Dimensions
object lengthlom—Dimensions
image areaAim²—Dimensions
object areaAom²—Dimensions
volume of them allVNm³—Dimensions
molar volumeVmm³—Dimensions
room to spareRV——Dimensions
θ=sr
arc over radius
Ω=Ar2
area over radius squared
dlight=c⋅tlight
distance at light speed
N=n⋅NA
entities in moles
θp=BD
baseline over distance
V=l3
volume of a cube
v=lt
average speed
a=vt
average acceleration
F=m⋅a
newtons second law
ρ=mV
mass per volume
W=F⋅l
force through a distance
S=6⋅l2
surface of a cube
V=4⋅π⋅r33
volume of a sphere
Sc=2⋅π⋅r⋅(r+h)
surface of a cylinder
A=l⋅b
area of a rectangle
Vs=l⋅b⋅h
volume from three sides
ρrel=ρρw
density against water
ml=lilo
lengths magnified
ml2=AiAo
areas magnified
VN=N⋅V
atoms side by side
RV=VmVN
molar over atomic volume