AutoMathic

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The Smart Calculator for Automated Math

(C) 1988, 2017  Kevin B. Belton


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We've all wondered about something, but never bothered to figure it out because it was "too much trouble".


"I could figure it out...", you say to yourself,


"but it would take too long", or
"I don't remember all the exact formulas", or
"I'd have to convert everything to the right units first".


"Besides", you continue, "it's not that important anyway..."


AutoMathic is intended to raise the bar of what you would consider "too much trouble".  The ability to conveniently get answers to mathematical questions frees you to ask more (and more interesting) questions since the program takes away the drudgery of getting answers!

AutoMathic is a flexible and very open-ended tool.  Like any tool it will enhance, not replace, your own ability!  To get the most out of AutoMathic, especially its "Converse" mode, you have to put some effort into it...  The more effort you put into understanding it, what it knows, and what it can do for you, the more you will get out of it!  AutoMathic will pay back that effort by allowing you to easily answer those questions that you could probably solve on your own (given time and/or patience), but wouldn't bother to for any number of reasons.  If only you could delegate to a smart assistant, leaving yourself more time to focus on more important things...


iOS7iPhonePortraitConverseGeneric_thumb.png If you have a mathematical question to answer, you can do it all yourself, or you can let AutoMathic help you do it!

With AutoMathic, you focus on specifying the problem, and let the computer do the mechanical work of solving it!


iOS7iPhoneLandscapeConverseGeneric_thumb.png
iOS7iPhonePortraitCommandGeneric_thumb.png





The documentation has lots of examples throughout the main body, in the quick-references, and in the lexicon...  The examples were chosen to demonstrate not only AutoMathic's vocabulary, but also a wide range of its grammar and default knowledge.  By seeing examples of its language in an understandable context, the user should begin to learn AutoMathic's "dialect" of English in the same way they might learn a foreign language by being immersed in it.

Here is just a sneak peek at the kinds of things understood by AutoMathic's "Converse" mode. 
These examples show how it doesn't just calculate...  It is smart enough to help you solve problems!  Just for fun, do some of these on your own to see how much work AutoMathic can save you!


* Except where noted, AutoMathic's output in these examples has been abbreviated to focus on how it:
  1. Understands your raw input, and
  2. Returns meaningful answers.
AutoMathic always generates full step-by-step solutions with as much or little detail as you want!



General Math

You enter...
AutoMathic understands...
Two
2
Two pair
4
-0.125
-0.125 (or -1 / 8)
A thousand
1,000
Seven trillion
7,000,000,000,000
Tens
10
100's
100
A sixteenth
0.0625 (or 1 / 16)
Two thirds
0.666666666666667 (or 2 / 3)
Four tenths
0.4 (or 2 / 5)
8 billionths
0.000000008
Three baker's dozens
39
6 gross
864
Pi
3.14159
e
2.71828
Golden ratio
1.61803
Avogadro's number
602,200,000,000,000,000,000,000


You enter...
AutoMathic understands...
Three Pi over two
4.712385
Twice the length added to twice the width
2 L + 2 W
Twice the sum of the length and the width
2 ( L + W )
3x the diameter
3 D
Half the mass
M / 2
A third of the sum of cost and tip
( C + T ) / 3
The ratio of width to height
W / H
The cost without half the taxes
C - T / 2


You enter...
AutoMathic understands...
Twice the sum of the width and the length is the perimeter. 2 (L + W) = P
The area would be half the base times the height.     B H
A = ---
     2

How much less than e is Pi over two?               3.14159
2.71828 - A = -------
                 2  

1.147485 is THE ANSWER.
Half of what age is three more than a fourth of itself? A       A
- = 3 + -
2       4

12 is THE AGE.


You enter...
AutoMathic understands...
12% of 275 million
12              
--- 275 (1000000)
100             

33,000,000 is 12 % OF 275 MILLION.
8.25 percent more than 198
8.25         
---- 198 + 198
100          

214.335 is 8.25 % MORE THAN 198.
12 pct less than 224.99
         3 (224.99)
224.99 - ----------
             25   

197.9912 is 12 % LESS THAN 224.99.
5% of each donation is overhead D    
-- = O
20   

How many millions is 12% of 275 million?
1000000 N = 33000000

33 is THE NUMBER.
What percentage of fifty is thirty?  P          
--- (50) = 30
100         

60 is THE PERCENTAGE.
Two dozen makes up only 20% of what total?          20  
12 (2) = --- T
         100 

         T
    24 = -
         5

120 is THE TOTAL.
What percent is one seventh?  A    1
--- = -
100   7

14.2857142857143 (or 100 / 7, or 14 & 2 / 7) is THE ANSWER.
Seventy percent is the ratio of successes to attempts. 7    S
-- = -
10   A
Thirty is a dozen less than 97% of what age?      97 A
30 = ---- - 12
     100

43.298969072165 is THE AGE.


You enter...
AutoMathic understands...
What percentage of a cost are its taxes? (Find P)

C P   

--- = T
100

7.62124711316397 is THE PERCENTAGE.
What must the price have been assuming that tips were 2.75? (Find P)

T = 2.75

16.9361046959199 is THE PRICE.
The price is 12% off of what when the cost is 225? (Find A)

        3 A
P = A - ---
        25

C = 225

236.195674994751 is THE AMOUNT.
The price consists of 0.65 per apple and 1.25 per drink. P = 0.65 A + 1.25 D
The price for each tire is 79.95.  Find the cost for 5 tires. P       
- = 79.95
T       

(Find 5C/T)

432.729375 is THE COST FOR 5 TIRES.
Find the price such that the cost is 225. (Find P)

C = 225

207.852193995381 is THE PRICE.


You enter...
AutoMathic understands...
The circumference is how many times the radius?
C = N R

C = 2 (3.14159) R

6.28318 is THE NUMBER.
21 is the difference between what circumference and what radius?
(Find R)

(Find C)

21 = C - R

C = 2 (3.14159) R

24.9748787661976 is THE CIRCUMFERENCE.

3.97487876619763 is THE RADIUS.
Compute the circumference where the diameter is 13.
(Find C)

D = 13

D = 2 R

C = 2 (3.14159) R

40.84067 is THE CIRCUMFERENCE.
When radius is 3, how much more than that's the circumference?
R = 3

(Find A)

A + R = C

C = 2 (3.14159) R

15.84954 is THE ANSWER.


Units and Unit Conversion

Category
Built-in Units (not including abbreviations)
Length
millimeters, centimeters, inches, feet, meters, yards, fathoms, furlongs, kilometers, miles, nautical miles, leagues, microns / micrometers, nanometers, angstroms, astronomical units, light years, parsecs
Time
nanoseconds, microseconds, milliseconds, seconds, minutes, hours, days, weeks, fortnights, months, years, decades, centuries, millenia
Weight & Mass
ounces, pounds, tons, milligrams, grams, kilograms, AMU, metric tons, kilotons
Liquid / Volume
teaspoons, tablespoons, fluid ounces, cups, pints, quarts, gallons, milliliters, liters, drops
Temperature
Fahrenheit, Celsius / Centigrade, Kelvin
Pressure
atmospheres, psi, torr, inches of mercury, millimeters of mercury, pascals, kilopascals, bars, millibars
Frequency
hertz, kilohertz, megahertz, gigahertz
Clock & Calendar Frequency
hourly, daily, weekly, monthly, quarterly, yearly, etc.
Angular
radians, degrees, grads, right angles, revolutions, rotations, circles, arcminutes, arcseconds
Information Storage
bits, nibbles, bytes, words, longwords, kilobytes, megabytes, gigabytes, terabytes, petabytes, kilobits, megabits, gigabits
Electrical
amperes, milliamperes, coulombs, volts, farads, microfarads, picofarads


You enter...
AutoMathic understands...
Inches per meter?
(Find I/M)

39.3700787401575 is INCHES PER METER.
Seconds in 1.5 hrs
(Find 1.5S/H)

5,400 is SECONDS IN 1.5 HRS.
What's 165[lb] in terms of [kg's]?
(Find 165K/P)

75 is 165 [ LB ] IN TERMS OF [ KG'S ].
How many unit cc's is 5 unit pints?
N   5
- = -
M   P

2,365.625 is THE NUMBER.
The temperature measured in degrees F is 78.
What's the temperature in terms of Kelvin?
F T = 78

(Find KT)

298.715555555556 is THE TEMPERATURE IN TERMS OF KELVIN.
Convert 90[PSI] into [mmHg].
(Find 90T/P)

4,653.0612244898 is CONVERT 90 [ PSI ] INTO [ MMHG ].
How many times 2.4[mHz] is 3.1[gHz]?
2.4 N   3.1
----- = ---
  M      G

1,291.66666666667 is THE NUMBER.
Find the cost daily when the cost semiannually is 475.
(Find C/D)

C     
- = 475
S    

2.6009582477755 is THE COST DAILY.
How many [deg's] is 1.5 pi [radians]?
N   1.5 (3.14159)
- = -------------
D         R     

270 is THE NUMBER.
What percentage of 54[mb] is 24[kb]?
27 P   24
---- = --
50 M   K

0.0434027777777778 (or 25 / 576) is THE PERCENTAGE.

Category
Component Units
Built-in Units and Constants (including some abbreviations)
Speed
length / time
miles per hour (MPH), kilometers per hour (KPH), feet per second (FPS), meters per second (MPS), knots, speed of sound (mach _), speed of light (_ C)
Angular Velocity
angle / time
revolutions per minute (RPM)
Acceleration
speed / time
G's, Gees, G forces, etc.
Momentum
mass x speed

Force
mass x acceleration newtons (N)
Density
mass / volume
density of water, mercury, alcohol, gasoline, air, helium, carbon dioxide, steam, aluminum, iron, steel, copper, lead, gold
Concentration
number / volume
moles per liter (_ molar)
Energy, Work, Torque
force x length
joules (J), kilojoules (kJ), calories (cal), food calories / kilocalories (kcal), watt hours (Wh), kilowatt hours (kWh), British thermal units (BTUs)
Power
energy / time
watts (W), kilowatts (kW), horsepower (HP)
Specific Heat
energy / mass
specific heat of water, ice, steam, mercury, alcohol, aluminum, copper, glass, iron, steel, lead, wood, flesh
Fuel Efficiency
length / volume
miles per gallon (MPG)


You enter...
AutoMathic understands...
The speed is 65[mph].
How many [kph] is a third of the speed?
    65 H
S = ----
     M 

H N   S
--- = -
 K    3

34.86912 is THE NUMBER.
What's 55[mph] in terms of [furlongs/fortnight]?
(Find 55FH/(Mf))

147,840 is 55 [ MPH ] IN TERMS OF [ FURLONGS / FORTNIGHT ].
3[drops/sec] converts to how many [gallons/day]?
3 S   N d
--- = ---
 D     G

3.42404227212682 is THE NUMBER.
Convert 1.3[gm/cc] into [lb/gal].
(Find 1.3MP/(Gg))

10.8251 is CONVERT 1.3 [ GM / CC ] INTO [ LB / GAL ].
How many times 5[km/l] is 23[mpg]?
5 L N   23 G
----- = ----
  K      M 

1.95587381770145 is THE NUMBER.
How many unit MB'S is 196[kilobits/sec] x 3.5[min]? N   686 S
- = -----
M    K m

4.90665435791016 is THE NUMBER.
There are 42 gallons a barrel.
How many [gallons/sec] does 35,000[barrels/day] convert to?
     G
42 = -
     B

N S   35000 D
--- = -------
 G       B  

17.0138888888889 is THE NUMBER.

Physics & Chemistry
You enter...
If the average speed is 130[mph], and the time is 73[sec], find the distance measured in kilometers.
AutoMathic understands...
   Let 'A' stand for "AVERAGE SPEED"

Let 'M' stand for "MILE"

Let 'H' stand for "HOUR"

So...
130 H
A = -----
M


Let 'T' stand for "TIME"

Let 'S' stand for "SECOND"

So...
73
T = --
S


Let 'D' stand for "DISTANCE"

Let 'K' stand for "KILOMETERS"

(Find DK)

Anything else?

> No.

I know that THERE ARE A HUNDRED CENTIMETERS IN A METER
I know that THERE ARE 2.54 CENTIMETERS PER INCH
I know that THERE ARE TWELVE INCHES IN A FOOT
I know that FOOT MEANS FEET
I know that THERE ARE 5280 FEET IN A MILE
I know that THERE ARE A THOUSAND METERS IN A KILOMETER
I know that THERE ARE SIXTY SECONDS IN A MINUTE
I know that THERE ARE SIXTY MINUTES IN AN HOUR
I know that DISTANCE IS AVERAGE SPEED TIMES TIME
I know that ACCELERATION IS SPEED OVER TIME
I know that DISTANCE IS ONE HALF SPEED TIMES TIME

Let 'C' stand for "CENTIMETERS"

Let 'm' stand for "METER"

So...
C
100 = -
m


Let 'I' stand for "INCH"

So...
C
2.54 = -
I


Let 'F' stand for "FOOT"

So...
I
12 = -
F


Let 'f' stand for "FEET"

So...
F = f


So...
f
5280 = -
M


So...
m
1000 = -
K


Let 'i' stand for "MINUTE"

So...
S
60 = -
i


So...
i
60 = -
H


So...
D = A T


Let 'a' stand for "ACCELERATION"

Let 's' stand for "SPEED"

So...
s
a = -
T


So...
T s
D = ---
2


Hypothetically...

M = 1

Solving for f:

f
5280 = -
M

f
- = 5280
M

1
f - = 5280
M

f = 5280 M

f = 5,280

F = f

F = 5,280

Solving for I:

I
12 = -
F

I
- = 12
F

1
I - = 12
F

I = 12 F

I = 63,360

Solving for C:

C
2.54 = -
I

C
- = 2.54
I

1
C - = 2.54
I

C = 2.54 I

C = 160,934.4

Solving for m:

C
100 = -
m

C
- = 100
m

C = 100 m

100 m = C

m (100) = C

C
m = ---
100

m = 1,609.344

Solving for K:

m
1000 = -
K

m
- = 1000
K

m = 1000 K

1000 K = m

K (1000) = m

m
K = ----
1000

K = 1.609344

Hypothetically...

i = 1

Solving for S:

S
60 = -
i

S
- = 60
i

1
S - = 60
i

S = 60 i

S = 60

Solving for H:

i
60 = -
H

i
- = 60
H

i = 60 H

60 H = i

H (60) = i

i
H = --
60

H = 0.0166666666666667 (or 1 / 60)

130 H
A = -----
M

A = 2.16666666666667 (or 13 / 6, or 2 & 1 / 6)

73
T = --
S

T = 1.21666666666667 (or 73 / 60, or 1 & 13 / 60)

D = A T

D = 2.63611111111111 (or 949 / 360, or 2 & 229 / 360)

Solving for s:

T s
D = ---
2

T s
--- = D
2

T
s - = D
2

T s = 2 D

s T = 2 D

2 D
s = ---
T

s = 4.33333333333333 (or 13 / 3, or 4 & 1 / 3)

s
a = -
T

a = 3.56164383561644 (or 260 / 73, or 3 & 41 / 73)

4.2424096 is THE DISTANCE MEASURED IN KILOMETERS.

You enter...
How many Gees is the acceleration if the speed was 60[mph], and the time was 3.4[sec]?
AutoMathic understands...
   Let 'N' stand for "NUMBER"

(Find N)

Let 'M' stand for "METERS"

Let 'S' stand for "SECOND"

Let 'A' stand for "ACCELERATION"

So...
9.8 N S S
--------- = A
M


Let 's' stand for "SPEED"

Let 'm' stand for "MILE"

Let 'H' stand for "HOUR"

So...
60 H
s = ----
m


Let 'T' stand for "TIME"

So...
3.4
T = ---
S


Anything else?

> no

I know that THERE ARE A HUNDRED CENTIMETERS IN A METER
I know that THERE ARE 2.54 CENTIMETERS PER INCH
I know that THERE ARE TWELVE INCHES IN A FOOT
I know that FOOT MEANS FEET
I know that THERE ARE 5280 FEET IN A MILE
I know that THERE ARE SIXTY SECONDS IN A MINUTE
I know that THERE ARE SIXTY MINUTES IN AN HOUR
I know that DISTANCE IS AVERAGE SPEED TIMES TIME
I know that ACCELERATION IS SPEED OVER TIME
I know that DISTANCE IS ONE HALF SPEED TIMES TIME

Let 'C' stand for "CENTIMETERS"

So...
C
100 = -
M


Let 'I' stand for "INCH"

So...
C
2.54 = -
I


Let 'F' stand for "FOOT"

So...
I
12 = -
F


Let 'f' stand for "FEET"

So...
F = f


So...
f
5280 = -
m


Let 'i' stand for "MINUTE"

So...
S
60 = -
i


So...
i
60 = -
H


Let 'D' stand for "DISTANCE"

Let 'a' stand for "AVERAGE SPEED"

So...
D = T a


So...
s
A = -
T


So...
T s
D = ---
2


Hypothetically...

M = 1

Solving for C:

C
100 = -
M

C
- = 100
M

1
C - = 100
M

C = 100 M

C = 100

Solving for I:

C
2.54 = -
I

C
- = 2.54
I

C = 2.54 I

2.54 I = C

I (2.54) = C

C
I = ----
2.54

I = 39.3700787401575

Solving for F:

I
12 = -
F

I
- = 12
F

I = 12 F

12 F = I

F (12) = I

I
F = --
12

F = 3.28083989501312

Solving for f:

F = f

f = F

f = 3.28083989501312

Solving for m:

f
5280 = -
m

f
- = 5280
m

f = 5280 m

5280 m = f

m (5280) = f

f
m = ----
5280

m = 0.000621371192237334

Hypothetically...

S = 1

3.4
T = ---
S

T = 3.4 (or 17 / 5, or 3 & 2 / 5)

Solving for i:

S
60 = -
i

S
- = 60
i

S = 60 i

60 i = S

i (60) = S

S
i = --
60

i = 0.0166666666666667 (or 1 / 60)

Solving for H:

i
60 = -
H

i
- = 60
H

i = 60 H

60 H = i

H (60) = i

i
H = --
60

H = 0.000277777777777778

60 H
s = ----
m

s = 26.8224

s
A = -
T

A = 7.88894117647059

T s
D = ---
2

D = 45.59808

Solving for N:

9.8 N S S
--------- = A
M

9.8 S S
N ------- = A
M

9.8 N S S = A M

N (9.8 S S) = A M

A M
N = -------
9.8 S S

N = 0.80499399759904

Solving for a:

D = T a

T a = D

a T = D

D
a = -
T

a = 13.4112

0.80499399759904 is THE NUMBER.

You enter...
Given that the distance is a quarter of a unit mile, and the time is 5.6[sec], find the speed measured in mph.
AutoMathic understands...
   Let 'D' stand for "DISTANCE"

Let 'M' stand for "MILE"

So...
1
D = ---
4 M


Let 'T' stand for "TIME"

Let 'S' stand for "SECOND"

So...
5.6
T = ---
S


Let 's' stand for "SPEED"

Let 'H' stand for "HOUR"

(Find Ms/H)

Anything else?

> No.

I know that THERE ARE SIXTY SECONDS IN A MINUTE
I know that THERE ARE SIXTY MINUTES IN AN HOUR
I know that DISTANCE IS AVERAGE SPEED TIMES TIME
I know that ACCELERATION IS SPEED OVER TIME
I know that DISTANCE IS ONE HALF SPEED TIMES TIME

Let 'm' stand for "MINUTE"

So...
S
60 = -
m


So...
m
60 = -
H


Let 'A' stand for "AVERAGE SPEED"

So...
D = A T


Let 'a' stand for "ACCELERATION"

So...
s
a = -
T


So...
T s
D = ---
2


Hypothetically...

H = 1

Solving for m:

m
60 = -
H

m
- = 60
H

1
m - = 60
H

m = 60 H

m = 60

Solving for S:

S
60 = -
m

S
- = 60
m

1
S - = 60
m

S = 60 m

S = 3,600

5.6
T = ---
S

T = 0.00155555555555556

(Setting the free variable 'D' to 1)

D = 1

Solving for M:

1
D = ---
4 M

1
--- = D
4 M

1 = 4 D M

4 D M = 1

M (4 D) = 1

1
M = ---
4 D

M = 0.25 (or 1 / 4)

Solving for A:

D = A T

A T = D

A T = D

D
A = -
T

A = 642.857142857143

Solving for s:

T s
D = ---
2

T s
--- = D
2

T
s - = D
2

T s = 2 D

s T = 2 D

2 D
s = ---
T

s = 1,285.71428571429

s
a = -
T

a = 826,530.612244898

321.428571428571 is THE SPEED MEASURED IN MPH.

You enter...
Find the mass measured in lbs so that the force is 750[N] when acceleration is 1 Gee.
AutoMathic understands...
   Let 'M' stand for "MASS"

Let 'P' stand for "POUND"

(Find MP)

Let 'F' stand for "FORCE"

Let 'K' stand for "KILOGRAM"

Let 'm' stand for "METERS"

Let 'S' stand for "SECOND"

So...
750 S S
F = -------
K m


Let 'A' stand for "ACCELERATION"

So...
9.8 S S
A = -------
m


Anything else?

> No.

I know that THERE ARE 2.2 POUNDS IN A KILOGRAM
I know that FORCE IS MASS TIMES ACCELERATION

So...
P
2.2 = -
K


So...
F = A M


Hypothetically...

m = 1

Hypothetically...

S = 1

9.8 S S
A = -------
m

A = 9.8 (or 49 / 5, or 9 & 4 / 5)

Hypothetically...

K = 1

750 S S
F = -------
K m

F = 750

Solving for P:

P
2.2 = -
K

P
- = 2.2
K

1
P - = 2.2
K

P = 2.2 K

P = 2.2 (or 11 / 5, or 2 & 1 / 5)

Solving for M:

F = A M

A M = F

M A = F

F
M = -
A

M = 76.530612244898

168.367346938776 is THE MASS MEASURED IN LBS.

You enter...
If the average speed is 50[kph], and the radius is 50[m], how many G's would the centripetal acceleration be?
AutoMathic understands...
   Let 'A' stand for "AVERAGE SPEED"

Let 'K' stand for "KILOMETER"

Let 'H' stand for "HOUR"

So...
50 H
A = ----
K


Let 'R' stand for "RADIUS"

Let 'M' stand for "METER"

So...
50
R = --
M


Let 'N' stand for "NUMBER"

(Find N)

Let 'S' stand for "SECOND"

Let 'C' stand for "CENTRIPETAL ACCELERATION"

So...
9.8 N S S
--------- = C
M


Anything else?

> No.

I know that THERE ARE A THOUSAND METERS IN A KILOMETER
I know that THERE ARE SIXTY SECONDS IN A MINUTE
I know that THERE ARE SIXTY MINUTES IN AN HOUR
I know that CENTRIPETAL ACCELERATION IS AVERAGE SPEED TIMES ITSELF OVER RADIUS

So...
M
1000 = -
K


Let 'm' stand for "MINUTE"

So...
S
60 = -
m


So...
m
60 = -
H


So...
A A
C = ---
R


Hypothetically...

K = 1

Solving for M:

M
1000 = -
K

M
- = 1000
K

1
M - = 1000
K

M = 1000 K

M = 1,000

50
R = --
M

R = 0.05 (or 1 / 20)

Hypothetically...

S = 1

Solving for m:

S
60 = -
m

S
- = 60
m

S = 60 m

60 m = S

m (60) = S

S
m = --
60

m = 0.0166666666666667 (or 1 / 60)

Solving for H:

m
60 = -
H

m
- = 60
H

m = 60 H

60 H = m

H (60) = m

m
H = --
60

H = 0.000277777777777778

50 H
A = ----
K

A = 0.0138888888888889 (or 1 / 72)

A A
C = ---
R

C = 0.00385802469135802

Solving for N:

9.8 N S S
--------- = C
M

9.8 S S
N ------- = C
M

9.8 N S S = C M

N (9.8 S S) = C M

C M
N = -------
9.8 S S

N = 0.393675988914084

0.393675988914084 is THE NUMBER.

You enter...
How many unit Joules is the work when the mass is 15[kg], acceleration is 1 G, and distance is 12[m]?
AutoMathic understands...
   Let 'N' stand for "NUMBER"

(Find N)

Let 'K' stand for "KILOGRAM"

Let 'M' stand for "METERS"

Let 'S' stand for "SECOND"

Let 'W' stand for "WORK"

So...
N S S
----- = W
K M M


Let 'm' stand for "MASS"

So...
15
m = --
K


Let 'A' stand for "ACCELERATION"

So...
9.8 S S
A = -------
M


Let 'D' stand for "DISTANCE"

So...
12
D = --
M


Anything else?

> no

I know that DISTANCE IS AVERAGE SPEED TIMES TIME
I know that ACCELERATION IS SPEED OVER TIME
I know that DISTANCE IS ONE HALF SPEED TIMES TIME
I know that FORCE IS MASS TIMES ACCELERATION
I know that WORK IS FORCE TIMES DISTANCE

Let 'a' stand for "AVERAGE SPEED"

Let 'T' stand for "TIME"

So...
D = T a


Let 's' stand for "SPEED"

So...
s
A = -
T


So...
T s
D = ---
2


Let 'F' stand for "FORCE"

So...
F = A m


So...
W = D F


(Setting the free variable 'S' to 1)

S = 1

(Setting the free variable 'K' to 1)

K = 1

15
m = --
K

m = 15

(Setting the free variable 'M' to 1)

M = 1

9.8 S S
A = -------
M

A = 9.8 (or 49 / 5, or 9 & 4 / 5)

12
D = --
M

D = 12

F = A m

F = 147

W = D F

W = 1,764

Solving for N:

N S S
----- = W
K M M

S S
N ----- = W
K M M

N S S = K M M W

N (S S) = K M M W

K M M W
N = -------
S S

N = 1,764

1,764 is THE NUMBER.

You enter...
How many unit Joules is the KE given that the average speed is 8.9[m/s], and the mass is 70[kg]?
AutoMathic understands...
   Let 'N' stand for "NUMBER"

(Find N)

Let 'K' stand for "KILOGRAM"

Let 'M' stand for "METERS"

Let 'S' stand for "SECOND"

Let 'k' stand for "KINETIC ENERGY"

So...
N S S
----- = k
K M M


Let 'A' stand for "AVERAGE SPEED"

So...
8.9 S
A = -----
M


Let 'm' stand for "MASS"

So...
70
m = --
K


Anything else?

> No.

I know that KINETIC ENERGY IS HALF MASS TIMES AVERAGE SPEED TIMES ITSELF

So...
A A m
k = -----
2


(Setting the free variable 'S' to 1)

S = 1

(Setting the free variable 'K' to 1)

K = 1

70
m = --
K

m = 70

(Setting the free variable 'M' to 1)

M = 1

8.9 S
A = -----
M

A = 8.9 (or 89 / 10, or 8 & 9 / 10)

A A m
k = -----
2

k = 2,772.35

Solving for N:

N S S
----- = k
K M M

S S
N ----- = k
K M M

N S S = K M M k

N (S S) = K M M k

K M M k
N = -------
S S

N = 2,772.35

2,772.35 is THE NUMBER.

You enter...
If the mass is 2,200[lb], and the height is 25[m], how many unit kJ's is the potential energy?
AutoMathic understands...
   Let 'M' stand for "MASS"

Let 'P' stand for "POUND"

So...
2200
M = ----
P


Let 'H' stand for "HEIGHT"

Let 'm' stand for "METER"

So...
25
H = --
m


Let 'N' stand for "NUMBER"

(Find N)

Let 'K' stand for "KILOGRAM"

Let 'S' stand for "SECOND"

Let 'p' stand for "POTENTIAL ENERGY"

So...
1000 N S S
---------- = p
K m m


Anything else?

> No.

I know that THERE ARE 2.2 POUNDS IN A KILOGRAM
I know that POTENTIAL ENERGY IS MASS TIMES 1 G TIMES HEIGHT

So...
P
2.2 = -
K


So...
9.8 H M S S
p = -----------
m


Hypothetically...

K = 1

Solving for P:

P
2.2 = -
K

P
- = 2.2
K

1
P - = 2.2
K

P = 2.2 K

P = 2.2 (or 11 / 5, or 2 & 1 / 5)

2200
M = ----
P

M = 1,000

(Setting the free variable 'H' to 1)

H = 1

Solving for m:

25
H = --
m

25
-- = H
m

25 = H m

H m = 25

m H = 25

25
m = --
H

m = 25

(Setting the free variable 'S' to 1)

S = 1

9.8 H M S S
p = -----------
m

p = 392

Solving for N:

1000 N S S
---------- = p
K m m

1000 S S
N -------- = p
K m m

1000 N S S = K m m p

N (1000 S S) = K m m p

K m m p
N = --------
1000 S S

N = 245

245 is THE NUMBER.

You enter...
How many unit kW would the power be assuming the energy is potential energy, mass is 160[lb], height was 4.5[m], and the time was 4[sec]?
AutoMathic understands...
   Let 'N' stand for "NUMBER"

(Find N)

Let 'K' stand for "KILOGRAM"

Let 'M' stand for "METERS"

Let 'S' stand for "SECOND"

Let 'P' stand for "POWER"

So...
1000 N S S S
------------ = P
K M M


Let 'E' stand for "ENERGY"

Let 'p' stand for "POTENTIAL ENERGY"

So...
E = p


Let 'm' stand for "MASS"

Let 'O' stand for "POUND"

So...
160
m = ---
O


Let 'H' stand for "HEIGHT"

So...
4.5
H = ---
M


Let 'T' stand for "TIME"

So...
4
T = -
S


Anything else?

> No.

I know that THERE ARE 2.2 POUNDS IN A KILOGRAM
I know that POTENTIAL ENERGY IS MASS TIMES 1 G TIMES HEIGHT
I know that POWER IS ENERGY OVER TIME

So...
O
2.2 = -
K


So...
9.8 H S S m
p = -----------
M


So...
E
P = -
T


Hypothetically...

K = 1

Solving for O:

O
2.2 = -
K

O
- = 2.2
K

1
O - = 2.2
K

O = 2.2 K

O = 2.2 (or 11 / 5, or 2 & 1 / 5)

160
m = ---
O

m = 72.7272727272727 (or 800 / 11, or 72 & 8 / 11)

(Setting the free variable 'S' to 1)

S = 1

4
T = -
S

T = 4

(Setting the free variable 'M' to 1)

M = 1

4.5
H = ---
M

H = 4.5 (or 9 / 2, or 4 & 1 / 2)

9.8 H S S m
p = -----------
M

p = 3,207.27272727273

E = p

E = 3,207.27272727273

E
P = -
T

P = 801.818181818182

Solving for N:

1000 N S S S
------------ = P
K M M

1000 S S S
N ---------- = P
K M M

1000 N S S S = K M M P

N (1000 S S S) = K M M P

K M M P
N = ----------
1000 S S S

N = 0.801818181818182 (or 441 / 550)

0.801818181818182 (or 441 / 550) is THE NUMBER.

You enter...
If the momentum was 2,500[lb] x 65[mph], what would the speed in [mph] be if the mass were only 200[lb]?
AutoMathic understands...
   Let 'M' stand for "MOMENTUM"

Let 'P' stand for "POUND"

Let 'm' stand for "MILE"

Let 'H' stand for "HOUR"

So...
162500 H
M = --------
P m


Let 'S' stand for "SPEED"

(Find Sm/H)

Let 'A' stand for "MASS"

So...
200
A = ---
P


Anything else?

> No.

I know that MOMENTUM IS MASS TIMES SPEED

So...
M = A S


Hypothetically...

H = 1

(Setting the free variable 'M' to 1)

M = 1

(Setting the free variable 'P' to 1)

P = 1

Solving for m:

162500 H
M = --------
P m

162500 H
-------- = M
P m

162500 H = M P m

M P m = 162500 H

m (M P) = 162500 H

162500 H
m = --------
M P

m = 162,500

200
A = ---
P

A = 200

Solving for S:

M = A S

A S = M

S A = M

M
S = -
A

S = 0.005 (or 1 / 200)

812.5 is THE SPEED IN [ MPH ].


You enter...
Calculate the torque in [N*m] when the radius is 18[cm], the mass is 132[lb], and acceleration is 1 G.
AutoMathic understands...
   Let 'T' stand for "TORQUE"

Let 'K' stand for "KILOGRAM"

Let 'M' stand for "METERS"

Let 'S' stand for "SECOND"

(Find KMMT/(SS))

Let 'R' stand for "RADIUS"

Let 'C' stand for "CENTIMETER"

So...
18
R = --
C


Let 'm' stand for "MASS"

Let 'P' stand for "POUND"

So...
132
m = ---
P


Let 'A' stand for "ACCELERATION"

So...
9.8 S S
A = -------
M


Anything else?

> No.

I know that THERE ARE A HUNDRED CENTIMETERS IN A METER
I know that THERE ARE 2.2 POUNDS IN A KILOGRAM
I know that FORCE IS MASS TIMES ACCELERATION
I know that TORQUE IS FORCE TIMES RADIUS

So...
C
100 = -
M


So...
P
2.2 = -
K


Let 'F' stand for "FORCE"

So...
F = A m


So...
T = F R


Hypothetically...

M = 1

Solving for C:

C
100 = -
M

C
- = 100
M

1
C - = 100
M

C = 100 M

C = 100

18
R = --
C

R = 0.18 (or 9 / 50)

Hypothetically...

S = 1

9.8 S S
A = -------
M

A = 9.8 (or 49 / 5, or 9 & 4 / 5)

Hypothetically...

K = 1

Solving for P:

P
2.2 = -
K

P
- = 2.2
K

1
P - = 2.2
K

P = 2.2 K

P = 2.2 (or 11 / 5, or 2 & 1 / 5)

132
m = ---
P

m = 60

F = A m

F = 588

T = F R

T = 105.84

105.84 is THE TORQUE IN [ N * M ].

You enter...
Find the spring constant in [N/m] given that the mass is 200[kg], and the displacement is 3[cm].  Acceleration is 1 G.
AutoMathic understands...
   Let 'S' stand for "SPRING CONSTANT"

Let 'K' stand for "KILOGRAM"

Let 'M' stand for "METERS"

Let 's' stand for "SECOND"

(Find KS/(ss))

Let 'm' stand for "MASS"

So...
200
m = ---
K


Let 'D' stand for "DISPLACEMENT"

Let 'C' stand for "CENTIMETER"

So...
3
D = -
C


Let 'A' stand for "ACCELERATION"

So...
9.8 s s
A = -------
M


Anything else?

> No.

I know that THERE ARE A HUNDRED CENTIMETERS IN A METER
I know that FORCE IS MASS TIMES ACCELERATION
I know that FORCE = SPRING CONSTANT X DISPLACEMENT

So...
C
100 = -
M


Let 'F' stand for "FORCE"

So...
F = A m


So...
F = D S


Hypothetically...

M = 1

Solving for C:

C
100 = -
M

C
- = 100
M

1
C - = 100
M

C = 100 M

C = 100

3
D = -
C

D = 0.03 (or 3 / 100)

Hypothetically...

s = 1

9.8 s s
A = -------
M

A = 9.8 (or 49 / 5, or 9 & 4 / 5)

(Setting the free variable 'm' to 1)

m = 1

Solving for K:

200
m = ---
K

200
--- = m
K

200 = K m

K m = 200

K m = 200

200
K = ---
m

K = 200

F = A m

F = 9.8 (or 49 / 5, or 9 & 4 / 5)

Solving for S:

F = D S

D S = F

S D = F

F
S = -
D

S = 326.666666666667 (or 980 / 3, or 326 & 2 / 3)

65,333.3333333333 is THE SPRING CONSTANT IN [ N / M ].

You enter...
If the wave speed is the speed of sound, and the frequency is F#, how many unit inches long is the wavelength?
AutoMathic understands...
   Let 'W' stand for "WAVE SPEED"

Let 'F' stand for "FEET"

Let 'S' stand for "SECOND"

So...
1088 S
W = ------
F


Let 'f' stand for "FREQUENCY"

Let 'H' stand for "HERTZ"

So...
370
f = ---
H


Let 'N' stand for "NUMBER"

(Find N)

Let 'I' stand for "INCHES"

Let 'w' stand for "WAVELENGTH"

So...
N
- = w
I


Anything else?

> No.

I know that THERE ARE TWELVE INCHES IN A FOOT
I know that FOOT MEANS FEET
I know that HERTZ MEANS 1 / SECOND
I know that WAVE SPEED IS WAVELENGTH TIMES FREQUENCY
I know that DOPPLER FREQUENCY = FREQUENCY X WAVE SPEED / ( WAVE SPEED + RELATIVE SPEED )
I know that DOPPLER SHIFT IS THE DIFFERENCE BETWEEN DOPPLER FREQUENCY AND FREQUENCY

Let 'O' stand for "FOOT"

So...
I
12 = -
O


So...
O = F


So...
1
H = -
S


So...
W = f w


Let 'D' stand for "DOPPLER FREQUENCY"

Let 'R' stand for "RELATIVE SPEED"

So...
W f
D = -----
R + W


Let 'd' stand for "DOPPLER SHIFT"

So...
d = D - f


Hypothetically...

S = 1

1
H = -
S

H = 1

370
f = ---
H

f = 370

Hypothetically...

O = 1

Solving for I:

I
12 = -
O

I
- = 12
O

1
I - = 12
O

I = 12 O

I = 12

Solving for F:

O = F

F = O

F = 1

1088 S
W = ------
F

W = 1,088

Solving for w:

W = f w

f w = W

w f = W

W
w = -
f

w = 2.94054054054054 (or 544 / 185, or 2 & 174 / 185)

Solving for N:

N
- = w
I

1
N - = w
I

N = I w

N = 35.2864864864865

35.2864864864865 is THE NUMBER.

You enter...
If the length is 36[inches], and the overtone is two, how many unit cm long is the wavelength?
AutoMathic understands...
   Let 'L' stand for "LENGTH"

Let 'I' stand for "INCHES"

So...
36
L = --
I


Let 'O' stand for "OVERTONE"

O = 2

O = 2

Let 'N' stand for "NUMBER"

(Find N)

Let 'C' stand for "CENTIMETER"

Let 'W' stand for "WAVELENGTH"

So...
N
- = W
C


Anything else?

> No.

I know that THERE ARE 2.54 CENTIMETERS PER INCH
I know that LENGTH IS THE HARMONIC TIMES WAVELENGTH OVER TWO
I know that THE OVERTONE IS ONE LESS THAN THE HARMONIC

So...
C
2.54 = -
I


Let 'H' stand for "HARMONIC"

So...
H W
L = ---
2


So...
O = H - 1


Solving for H:

O = - 1 + H

- 1 + H = O

H = 1 + O

H = 3

Hypothetically...

I = 1

36
L = --
I

L = 36

Solving for C:

C
2.54 = -
I

C
- = 2.54
I

1
C - = 2.54
I

C = 2.54 I

C = 2.54 (or 127 / 50, or 2 & 27 / 50)

Solving for W:

H W
L = ---
2

H W
--- = L
2

H
W - = L
2

H W = 2 L

W H = 2 L

2 L
W = ---
H

W = 24

Solving for N:

N
- = W
C

1
N - = W
C

N = C W

N = 60.96

60.96 is THE NUMBER.

You enter...
Find the doppler shift in unit Hz.  The wave speed is mach 1, the frequency is G natural, and the relative speed is 45[mph].
AutoMathic understands...
   Let 'D' stand for "DOPPLER SHIFT"

Let 'H' stand for "HERTZ"

(Find DH)

Let 'W' stand for "WAVE SPEED"

Let 'F' stand for "FEET"

Let 'S' stand for "SECOND"

So...
1088 S
W = ------
F


Let 'f' stand for "FREQUENCY"

So...
392
f = ---
H


Let 'R' stand for "RELATIVE SPEED"

Let 'M' stand for "MILE"

Let 'h' stand for "HOUR"

So...
45 h
R = ----
M


Anything else?

> No.

I know that THERE ARE 5280 FEET IN A MILE
I know that THERE ARE SIXTY SECONDS IN A MINUTE
I know that THERE ARE SIXTY MINUTES IN AN HOUR
I know that HERTZ MEANS 1 / SECOND
I know that WAVE SPEED IS WAVELENGTH TIMES FREQUENCY
I know that DOPPLER FREQUENCY = FREQUENCY X WAVE SPEED / ( WAVE SPEED + RELATIVE SPEED )
I know that DOPPLER SHIFT IS THE DIFFERENCE BETWEEN DOPPLER FREQUENCY AND FREQUENCY

So...
F
5280 = -
M


Let 'm' stand for "MINUTE"

So...
S
60 = -
m


So...
m
60 = -
h


So...
1
H = -
S


Let 'w' stand for "WAVELENGTH"

So...
W = f w


Let 'd' stand for "DOPPLER FREQUENCY"

So...
W f
d = -----
R + W


So...
D = d - f


Hypothetically...

S = 1

Solving for m:

S
60 = -
m

S
- = 60
m

S = 60 m

60 m = S

m (60) = S

S
m = --
60

m = 0.0166666666666667 (or 1 / 60)

Solving for h:

m
60 = -
h

m
- = 60
h

m = 60 h

60 h = m

h (60) = m

m
h = --
60

h = 0.000277777777777778

1
H = -
S

H = 1

392
f = ---
H

f = 392

Hypothetically...

M = 1

45 h
R = ----
M

R = 0.0125 (or 1 / 80)

Solving for F:

F
5280 = -
M

F
- = 5280
M

1
F - = 5280
M

F = 5280 M

F = 5,280

1088 S
W = ------
F

W = 0.206060606060606 (or 34 / 165)

Solving for w:

W = f w

f w = W

w f = W

W
w = -
f

w = 0.000525664811379097

W f
d = -----
R + W

d = 369.580589254766

D = d - f

D = -22.419410745234

-22.419410745234 is THE DOPPLER SHIFT IN UNIT HZ.

You enter...
How many trillionths of a unit Coulomb is the charge when the capacitance is 53[pF], and the voltage is 12[V]?
AutoMathic understands...
   Let 'N' stand for "NUMBER"

(Find N)

Let 'C' stand for "COULOMB"

Let 'c' stand for "CHARGE"

So...
N
--------------- = c
1000000000000 C


Let 'A' stand for "CAPACITANCE"

Let 'P' stand for "PICOFARAD"

So...
53
A = --
P


Let 'V' stand for "VOLTAGE"

Let 'v' stand for "VOLT"

So...
12
V = --
v


Anything else?

> No.

I know that VOLTS = JOULES PER COULOMB
I know that FARADS = COULOMBS PER VOLT
I know that THERE ARE A TRILLION PICOFARADS IN A FARAD
I know that CAPACITANCE IS CHARGE OVER VOLTAGE
I know that ELECTRICAL ENERGY IS CHARGE TIMES VOLTAGE

Let 'K' stand for "KILOGRAM"

Let 'M' stand for "METERS"

Let 'S' stand for "SECOND"

So...
C K M M
v = -------
S S


Let 'F' stand for "FARADS"

So...
C
F = -
v


So...
P
1000000000000 = -
F


So...
c
A = -
V


Let 'E' stand for "ELECTRICAL ENERGY"

So...
E = V c


Hypothetically...

v = 1

12
V = --
v

V = 12

Hypothetically...

F = 1

Solving for C:

C
F = -
v

C
- = F
v

1
C - = F
v

C = F v

C = 1

Solving for P:

P
1000000000000 = -
F

P
- = 1000000000000
F

1
P - = 1000000000000
F

P = 1000000000000 F

P = 1,000,000,000,000

53
A = --
P

A = 0.000000000053

Solving for c:

c
A = -
V

c
- = A
V

1
c - = A
V

c = A V

c = 0.000000000636

E = V c

E = 0.000000007632

Solving for N:

N
--------------- = c
1000000000000 C

1
N --------------- = c
1000000000000 C

N = 1000000000000 C c

N = 636

636 is THE NUMBER.

You enter...
Calculate the charge measured in Coulombs when the current is 2.5[amps], and the time is 4[min].
AutoMathic understands...
   Let 'C' stand for "CHARGE"

Let 'c' stand for "COULOMBS"

(Find Cc)

Let 'U' stand for "CURRENT"

Let 'A' stand for "AMPERE"

So...
2.5
U = ---
A


Let 'T' stand for "TIME"

Let 'M' stand for "MINUTE"

So...
4
T = -
M


Anything else?

> No.

I know that THERE ARE SIXTY SECONDS IN A MINUTE
I know that AMPERES = COULOMBS PER SECOND
I know that CURRENT IS CHARGE OVER TIME

Let 'S' stand for "SECONDS"

So...
S
60 = -
M


So...
c
A = -
S


So...
C
U = -
T


Hypothetically...

M = 1

4
T = -
M

T = 4

Solving for S:

S
60 = -
M

S
- = 60
M

1
S - = 60
M

S = 60 M

S = 60

(Setting the free variable 'U' to 1)

U = 1

Solving for A:

2.5
U = ---
A

2.5
--- = U
A

2.5 = A U

A U = 2.5

A U = 2.5

2.5
A = ---
U

A = 2.5 (or 5 / 2, or 2 & 1 / 2)

Solving for c:

c
A = -
S

c
- = A
S

1
c - = A
S

c = A S

c = 150

Solving for C:

C
U = -
T

C
- = U
T

1
C - = U
T

C = T U

C = 4

600 is THE CHARGE MEASURED IN COULOMBS.

You enter...
Given that voltage is 6[V], and the current is 300[mA], find the resistance measured in ohms.
AutoMathic understands...
   Let 'V' stand for "VOLTAGE"

Let 'v' stand for "VOLT"

So...
6
V = -
v


Let 'C' stand for "CURRENT"

Let 'M' stand for "MILLIAMPERE"

So...
300
C = ---
M


Let 'R' stand for "RESISTANCE"

Let 'O' stand for "OHMS"

(Find OR)

Anything else?

> No.

I know that THERE ARE A THOUSAND MILLIAMPERES IN AN AMPERE
I know that OHMS = VOLTS / AMPERES
I know that VOLTAGE IS CURRENT TIMES RESISTANCE

Let 'A' stand for "AMPERE"

So...
M
1000 = -
A


So...
v
O = -
A


So...
V = C R


Hypothetically...

A = 1

Solving for M:

M
1000 = -
A

M
- = 1000
A

1
M - = 1000
A

M = 1000 A

M = 1,000

300
C = ---
M

C = 0.3 (or 3 / 10)

(Setting the free variable 'V' to 1)

V = 1

Solving for v:

6
V = -
v

6
- = V
v

6 = V v

V v = 6

v V = 6

6
v = -
V

v = 6

v
O = -
A

O = 6

Solving for R:

V = C R

C R = V

R C = V

V
R = -
C

R = 3.33333333333333 (or 10 / 3, or 3 & 1 / 3)

20 is THE RESISTANCE MEASURED IN OHMS.

You enter...
If the power is 40[W], and the voltage is 12[V], find the resistance measured in ohms, and the current measured in amps.
AutoMathic understands...
   Let 'P' stand for "POWER"

Let 'K' stand for "KILOGRAM"

Let 'M' stand for "METERS"

Let 'S' stand for "SECOND"

So...
40 S S S
P = --------
K M M


Let 'V' stand for "VOLTAGE"

Let 'v' stand for "VOLT"

So...
12
V = --
v


Let 'R' stand for "RESISTANCE"

Let 'O' stand for "OHMS"

(Find OR)

Let 'C' stand for "CURRENT"

Let 'A' stand for "AMPERE"

(Find AC)

Anything else?

> No.

I know that AMPERES = COULOMBS PER SECOND
I know that VOLTS = JOULES PER COULOMB
I know that OHMS = VOLTS / AMPERES
I know that FARADS = COULOMBS PER VOLT
I know that VOLTAGE IS CURRENT TIMES RESISTANCE
I know that POWER IS CURRENT TIMES VOLTAGE. ASSUME KILOGRAMS=1; METERS=1

Let 'c' stand for "COULOMBS"

So...
c
A = -
S


So...
K M M c
v = -------
S S


So...
v
O = -
A


Let 'F' stand for "FARADS"

So...
c
F = -
v


So...
V = C R


So...
P = C V


K = 1

K = 1

M = 1

M = 1

Hypothetically...

S = 1

40 S S S
P = --------
K M M

P = 40

Hypothetically...

v = 1

12
V = --
v

V = 12

Solving for c:

K M M c
v = -------
S S

K M M c
------- = v
S S

K M M
c ----- = v
S S

K M M c = S S v

c (K M M) = S S v

S S v
c = -----
K M M

c = 1

c
F = -
v

F = 1

Solving for C:

P = C V

C V = P

C V = P

P
C = -
V

C = 3.33333333333333 (or 10 / 3, or 3 & 1 / 3)

c
A = -
S

A = 1

v
O = -
A

O = 1

Solving for R:

V = C R

C R = V

R C = V

V
R = -
C

R = 3.6 (or 18 / 5, or 3 & 3 / 5)

3.6 (or 18 / 5, or 3 & 3 / 5) is THE RESISTANCE MEASURED IN OHMS.

3.33333333333333 (or 10 / 3, or 3 & 1 / 3) is THE CURRENT MEASURED IN AMPS.

You enter...
If the current is 15[Amps] when the voltage is 120[V], find the electrical energy in [kWh] when the time is 90[hr's].
AutoMathic understands...
   Let 'C' stand for "CURRENT"

Let 'A' stand for "AMPERE"

So...
15
C = --
A


Let 'V' stand for "VOLTAGE"

Let 'v' stand for "VOLT"

So...
120
V = ---
v


Let 'E' stand for "ELECTRICAL ENERGY"

Let 'K' stand for "KILOGRAM"

Let 'M' stand for "METERS"

Let 'S' stand for "SECOND"

Let 'H' stand for "HOUR"

(Find EHKMM/(1000SSS))

Let 'T' stand for "TIME"

So...
90
T = --
H


Anything else?

> No.

I know that THERE ARE SIXTY SECONDS IN A MINUTE
I know that THERE ARE SIXTY MINUTES IN AN HOUR
I know that POWER IS ENERGY OVER TIME
I know that AMPERES = COULOMBS PER SECOND
I know that VOLTS = JOULES / COULOMB
I know that OHMS = VOLTS / AMPERES
I know that FARADS = COULOMBS PER VOLT
I know that CURRENT IS CHARGE OVER TIME
I know that VOLTAGE IS CURRENT TIMES RESISTANCE
I know that CAPACITANCE IS CHARGE OVER VOLTAGE
I know that ELECTRICAL ENERGY IS CHARGE TIMES VOLTAGE
I know that POWER IS CURRENT TIMES VOLTAGE. ASSUME KILOGRAMS=1; METERS=1

Let 'm' stand for "MINUTE"

So...
S
60 = -
m


So...
m
60 = -
H


Let 'P' stand for "POWER"

Let 'e' stand for "ENERGY"

So...
e
P = -
T


Let 'c' stand for "COULOMBS"

So...
c
A = -
S


So...
K M M
v = -----
S S c


Let 'O' stand for "OHMS"

So...
v
O = -
A


Let 'F' stand for "FARADS"

So...
c
F = -
v


Let 'h' stand for "CHARGE"

So...
h
C = -
T


Let 'R' stand for "RESISTANCE"

So...
V = C R


Let 'a' stand for "CAPACITANCE"

So...
h
a = -
V


So...
E = V h


So...
P = C V


K = 1

K = 1

M = 1

M = 1

Hypothetically...

m = 1

Solving for S:

S
60 = -
m

S
- = 60
m

1
S - = 60
m

S = 60 m

S = 60

Solving for H:

m
60 = -
H

m
- = 60
H

m = 60 H

60 H = m

H (60) = m

m
H = --
60

H = 0.0166666666666667 (or 1 / 60)

90
T = --
H

T = 5,400

Hypothetically...

v = 1

120
V = ---
v

V = 120

Solving for c:

K M M
v = -----
S S c

K M M
----- = v
S S c

K M M = S S c v

S S c v = K M M

c (S S v) = K M M

K M M
c = -----
S S v

c = 0.000277777777777778

c
F = -
v

F = 0.000277777777777778

c
A = -
S

A = 0.00000462962962962963

v
O = -
A

O = 216,000

15
C = --
A

C = 3,240,000

Solving for h:

h
C = -
T

h
- = C
T

1
h - = C
T

h = C T

h = 17,496,000,000

Solving for R:

V = C R

C R = V

R C = V

V
R = -
C

R = 0.000037037037037037

h
a = -
V

a = 145,800,000

E = V h

E = 2,099,520,000,000

P = C V

P = 388,800,000

Solving for e:

e
P = -
T

e
- = P
T

1
e - = P
T

e = P T

e = 2,099,520,000,000

162 is THE ELECTRICAL ENERGY IN [ KWH ].

You enter...
Find the mass per particle measured in amu, when the mass for 6 moles is 84[gm].
AutoMathic understands...
   Let 'M' stand for "MASS"

Let 'P' stand for "PARTICLE"

Let 'A' stand for "AMU"

(Find AM/P)

Let 'm' stand for "MOLES"

Let 'G' stand for "GRAM"

So...
6 M 84
--- = --
m G


Anything else?

> No.

I know that AVOGADRO'S NUMBER IS THE NUMBER OF PARTICLES PER MOLE
I know that MOLES IS GRAMS / FORMULA WEIGHT
I know that AVOGADRO'S NUMBER IS THE NUMBER OF AMU PER GRAM

So...
P
602200000000000000000000 = -
m


Let 'F' stand for "FORMULA WEIGHT"

So...
G
m = -
F


So...
A
602200000000000000000000 = -
G


Hypothetically...

P = 1

Solving for m:

P
602200000000000000000000 = -
m

P
- = 602200000000000000000000
m

P = 602200000000000000000000 m

602200000000000000000000 m = P

m (602200000000000000000000) = P

P
m = ------------------------
602200000000000000000000

m = 0.00000000000000000000000166057788110262

Hypothetically...

G = 1

Solving for M:

6 M 84
--- = --
m G

6 84
M - = --
m G

84 m
6 M = ----
G

84 m
M (6) = ----
G

14 m
M = ----
G

M = 0.0000000000000000000000232480903354367

Solving for F:

G
m = -
F

G
- = m
F

G = F m

F m = G

F m = G

G
F = -
m

F = 602,200,000,000,000,000,000,000

Solving for A:

A
602200000000000000000000 = -
G

A
- = 602200000000000000000000
G

1
A - = 602200000000000000000000
G

A = 602200000000000000000000 G

A = 602,200,000,000,000,000,000,000

14 is THE MASS PER PARTICLE MEASURED IN AMU.

You enter...
If the formula weight is 106, find the number of grams such that the molar concentration is 4 when the number of ml is 1,500.
AutoMathic understands...
   Let 'F' stand for "FORMULA WEIGHT"

F = 106

F = 106

Let 'G' stand for "GRAMS"

(Find G)

Let 'M' stand for "MOLARITY"

M = 4

M = 4

Let 'm' stand for "MILLILITER"

m = 1500

m = 1,500

Anything else?

> No.

I know that THERE ARE A THOUSAND MILLILITERS IN A LITER
I know that MOLARITY IS MOLES PER LITER
I know that MOLES IS GRAMS / FORMULA WEIGHT
I know that SPECIFIC GRAVITY = DENSITY / (1 [GRAM/ML])
I know that PRESSURE IN [ATM] X VOLUME IN [L] = MOLES X 0.0821 X TEMPERATURE IN [KELVIN]
I know that HEAT ENERGY IN [JOULES] IS 3/2 X MOLES X 8.314 X TEMPERATURE IN [KELVIN]. ASSUME SECONDS=1; METERS=1

Let 'L' stand for "LITER"

So...
m
1000 = -
L


Solving for L:

m
1000 = -
L

m
- = 1000
L

m = 1000 L

1000 L = m

L (1000) = m

m
L = ----
1000

L = 1.5 (or 3 / 2, or 1 & 1 / 2)

Let 'O' stand for "MOLES"

So...
O
M = -
L


Solving for O:

O
M = -
L

O
- = M
L

1
O - = M
L

O = L M

O = 6

So...
G
O = -
F


Solving for G:

G
O = -
F

G
- = O
F

1
G - = O
F

G = F O

G = 636

Let 'S' stand for "SPECIFIC GRAVITY"

Let 'D' stand for "DENSITY"

So...
D G
S = ---
m


Let 'P' stand for "PRESSURE"

Let 'A' stand for "ATMOSPHERE"

Let 'V' stand for "VOLUME"

Let 'T' stand for "TEMPERATURE"

Let 'K' stand for "KELVIN"

So...
A L P V = 0.0821 K O T


Let 'H' stand for "HEAT ENERGY"

Let 'k' stand for "KILOGRAM"

Let 'E' stand for "METERS"

Let 's' stand for "SECOND"

So...
E E H k 3 (8.314) K O T
------- = ---------------
s s 2


s = 1

s = 1

E = 1

E = 1

636 is THE NUMBER OF GRAMS.

You enter...
If the density is half the density of steel, and 75% of the volume is 2[gal], find the mass measured in pounds.
AutoMathic understands...
   Let 'D' stand for "DENSITY"

Let 'G' stand for "GRAM"

Let 'M' stand for "MILLILITER"

So...
7.8 M
D = -----
2 G


Let 'V' stand for "VOLUME"

Let 'g' stand for "GALLON"

So...
3 V 2
--- = -
4 g


Let 'm' stand for "MASS"

Let 'P' stand for "POUNDS"

(Find Pm)

Anything else?

> No.

I know that THERE ARE 2.2 POUNDS IN A KILOGRAM
I know that THERE ARE A THOUSAND GRAMS IN A KILOGRAM
I know that THERE ARE 3.785 LITERS IN A GALLON
I know that THERE ARE A THOUSAND MILLILITERS IN A LITER
I know that DENSITY IS THE RATIO OF MASS TO VOLUME
I know that SPECIFIC GRAVITY = DENSITY / (1 [GRAM/ML])

Let 'K' stand for "KILOGRAM"

So...
P
2.2 = -
K


So...
G
1000 = -
K


Let 'L' stand for "LITERS"

So...
L
3.785 = -
g


So...
M
1000 = -
L


So...
m
D = -
V


Let 'S' stand for "SPECIFIC GRAVITY"

So...
D G
S = ---
M


Hypothetically...

M = 1

Solving for L:

M
1000 = -
L

M
- = 1000
L

M = 1000 L

1000 L = M

L (1000) = M

M
L = ----
1000

L = 0.001

Solving for g:

L
3.785 = -
g

L
- = 3.785
g

L = 3.785 g

3.785 g = L

g (3.785) = L

L
g = -----
3.785

g = 0.000264200792602378

Solving for V:

3 V 2
--- = -
4 g

3 2
V - = -
4 g

8
3 V = -
g

8
V (3) = -
g

8
V = ---
3 g

V = 10,093.3333333333

Hypothetically...

K = 1

Solving for P:

P
2.2 = -
K

P
- = 2.2
K

1
P - = 2.2
K

P = 2.2 K

P = 2.2 (or 11 / 5, or 2 & 1 / 5)

Solving for G:

G
1000 = -
K

G
- = 1000
K

1
G - = 1000
K

G = 1000 K

G = 1,000

7.8 M
D = -----
2 G

D = 0.0039

Solving for m:

m
D = -
V

m
- = D
V

1
m - = D
V

m = D V

m = 39.364

D G
S = ---
M

S = 3.9 (or 39 / 10, or 3 & 9 / 10)

86.6008 is THE MASS MEASURED IN POUNDS.

You enter...
If the specific gravity is 0.668, find the density measured in lb/gal.
AutoMathic understands...
   Let 'S' stand for "SPECIFIC GRAVITY"

S = 0.668

S = 0.668 (or 167 / 250)

Let 'D' stand for "DENSITY"

Let 'P' stand for "POUND"

Let 'G' stand for "GALLON"

(Find DP/G)

Anything else?

> No.

I know that THERE ARE 2.2 POUNDS IN A KILOGRAM
I know that THERE ARE A THOUSAND GRAMS IN A KILOGRAM
I know that THERE ARE 3.785 LITERS IN A GALLON
I know that THERE ARE A THOUSAND MILLILITERS IN A LITER
I know that SPECIFIC GRAVITY = DENSITY / (1 [GRAM/ML])

Let 'K' stand for "KILOGRAM"

So...
P
2.2 = -
K


Let 'g' stand for "GRAMS"

So...
g
1000 = -
K


Let 'L' stand for "LITERS"

So...
L
3.785 = -
G


Let 'M' stand for "MILLILITERS"

So...
M
1000 = -
L


So...
D g
S = ---
M


Hypothetically...

K = 1

Solving for P:

P
2.2 = -
K

P
- = 2.2
K

1
P - = 2.2
K

P = 2.2 K

P = 2.2 (or 11 / 5, or 2 & 1 / 5)

Solving for g:

g
1000 = -
K

g
- = 1000
K

1
g - = 1000
K

g = 1000 K

g = 1,000

Hypothetically...

G = 1

Solving for L:

L
3.785 = -
G

L
- = 3.785
G

1
L - = 3.785
G

L = 3.785 G

L = 3.785 (or 757 / 200, or 3 & 157 / 200)

Solving for M:

M
1000 = -
L

M
- = 1000
L

1
M - = 1000
L

M = 1000 L

M = 3,785

Solving for D:

D g
S = ---
M

D g
--- = S
M

g
D - = S
M

D g = M S

D g = M S

M S
D = ---
g

D = 2.52838

5.562436 is THE DENSITY MEASURED IN LB / GAL.

You enter...
Find the number of nucleons, and the atomic weight.  The atomic number is 6, the ratio of protons to neutrons is 1:1, and the ionic charge is +2.
AutoMathic understands...
   Let 'N' stand for "NUCLEONS"

(Find N)

Let 'F' stand for "FORMULA WEIGHT"

(Find F)

Let 'A' stand for "ATOMIC NUMBER"

A = 6

A = 6

Let 'P' stand for "PROTONS"

Let 'n' stand for "NEUTRONS"

So...
P
- = 1
n


Let 'I' stand for "IONIC NUMBER"

I = 2

I = 2

Anything else?

> No.

I know that ATOMIC NUMBER IS THE NUMBER OF PROTONS IN AN ATOM
I know that MASS NUMBER IS THE SUM OF THE ATOMIC NUMBER AND NEUTRONS IN THE ATOM
I know that MASS NUMBER IS THE NUMBER OF NUCLEONS
I know that IONIC NUMBER IS THE DIFFERENCE BETWEEN THE NUMBER OF PROTONS PER ATOM AND ELECTRONS PER ATOM
I know that ATOMIC WEIGHT X ATOMS = 1.007276 X PROTONS + 1.008665 X NEUTRONS + 0.0005486 X ELECTRONS

Let 'a' stand for "ATOM"

So...
P
A = -
a


Let 'M' stand for "MASS NUMBER"

So...
n
M = A + -
a


So...
M = N


Let 'E' stand for "ELECTRONS"

So...
P - E
I = -----
a


So...
F a = 0.0005486 E + 1.007276 P + 1.008665 n


Hypothetically...

a = 1

Solving for P:

P
A = -
a

P
- = A
a

1
P - = A
a

P = A a

P = 6

Solving for E:

P - E
I = -----
a

P - E
----- = I
a

P E
- - - = I
a a

E P
- - = I - -
a a

/ 1\ P
E ( - - ) = I - -
\ a/ a

E P
- = - - I
a a

1 P
E - = - - I
a a

E = P - I a

E = 4

Solving for n:

P
- = 1
n

P = n

n = P

n = 6

n
M = A + -
a

M = 12

Solving for N:

M = N

N = M

N = 12

Solving for F:

F a = 0.0005486 E + 1.007276 P + 1.008665 n

F a = 0.0005486 E + 1.007276 P + 1.008665 n

0.0005486 E + 1.007276 P + 1.008665 n
F = -------------------------------------
a

F = 12.0978404

12 is THE NUMBER OF NUCLEONS.

12.0978404 is THE ATOMIC WEIGHT.

You enter...
Compute the volume measured in liters, if the number of moles is 1, pressure is standard pressure, and temperature is standard temperature.
AutoMathic understands...
   Let 'V' stand for "VOLUME"

Let 'L' stand for "LITERS"

(Find LV)

Let 'M' stand for "MOLES"

M = 1

M = 1

Let 'P' stand for "PRESSURE"

Let 'A' stand for "ATMOSPHERE"

So...
1
P = -
A


Let 'T' stand for "TEMPERATURE"

Let 'K' stand for "KELVIN"

So...
273.159
T = -------
K


Anything else?

> No.

I know that MOLARITY IS MOLES PER LITER
I know that PRESSURE IN [ATM] X VOLUME IN [L] = MOLES X 0.0821 X TEMPERATURE IN [KELVIN]
I know that HEAT ENERGY IN [JOULES] IS 3/2 X MOLES X 8.314 X TEMPERATURE IN [KELVIN]. ASSUME SECONDS=1; METERS=1

Let 'm' stand for "MOLARITY"

So...
M
m = -
L


So...
A L P V = 0.0821 K M T


Let 'H' stand for "HEAT ENERGY"

Let 'k' stand for "KILOGRAM"

Let 'E' stand for "METERS"

Let 'S' stand for "SECOND"

So...
E E H k 3 (8.314) K M T
------- = ---------------
S S 2


S = 1

S = 1

E = 1

E = 1

Hypothetically...

L = 1

M
m = -
L

m = 1

(Setting the free variable 'P' to 1)

P = 1

Solving for A:

1
P = -
A

1
- = P
A

1 = A P

A P = 1

A P = 1

1
A = -
P

A = 1

(Setting the free variable 'T' to 1)

T = 1

Solving for K:

273.159
T = -------
K

273.159
------- = T
K

273.159 = K T

K T = 273.159

K T = 273.159

273.159
K = -------
T

K = 273.159

Solving for V:

A L P V = 0.0821 K M T

V (A L P) = 0.0821 K M T

0.0821 K M T
V = ------------
A L P

V = 22.4263539

22.4263539 is THE VOLUME MEASURED IN LITERS.

You enter...
Calculate the heat energy in [J] when the number of moles is 5, and the temperature is room temperature.
AutoMathic understands...
   Let 'H' stand for "HEAT ENERGY"

Let 'K' stand for "KILOGRAM"

Let 'M' stand for "METERS"

Let 'S' stand for "SECOND"

(Find HKMM/(SS))

Let 'm' stand for "MOLES"

m = 5

m = 5

Let 'T' stand for "TEMPERATURE"

Let 'k' stand for "KELVIN"

So...
293
T = ---
k


Anything else?

> No.

I know that THERE ARE A THOUSAND GRAMS IN A KILOGRAM
I know that MOLARITY IS MOLES PER LITER
I know that MOLES IS GRAMS / FORMULA WEIGHT
I know that PRESSURE IN [ATM] X VOLUME IN [L] = MOLES X 0.0821 X TEMPERATURE IN [KELVIN]
I know that HEAT ENERGY IN [JOULES] IS 3/2 X MOLES X 8.314 X TEMPERATURE IN [KELVIN]. ASSUME SECONDS=1; METERS=1

Let 'G' stand for "GRAMS"

So...
G
1000 = -
K


Let 'O' stand for "MOLARITY"

Let 'L' stand for "LITER"

So...
m
O = -
L


Let 'F' stand for "FORMULA WEIGHT"

So...
G
m = -
F


Let 'P' stand for "PRESSURE"

Let 'A' stand for "ATMOSPHERE"

Let 'V' stand for "VOLUME"

So...
A L P V = 0.0821 T k m


So...
H K M M 3 (8.314) T k m
------- = ---------------
S S 2


S = 1

S = 1

M = 1

M = 1

Hypothetically...

K = 1

Solving for G:

G
1000 = -
K

G
- = 1000
K

1
G - = 1000
K

G = 1000 K

G = 1,000

Solving for F:

G
m = -
F

G
- = m
F

G = F m

F m = G

F m = G

G
F = -
m

F = 200

Hypothetically...

L = 1

m
O = -
L

O = 5

(Setting the free variable 'T' to 1)

T = 1

Solving for k:

293
T = ---
k

293
--- = T
k

293 = T k

T k = 293

k T = 293

293
k = ---
T

k = 293

Solving for H:

H K M M 3 (8.314) T k m
------- = ---------------
S S 2

K M M 3 (8.314) T k m
H ----- = ---------------
S S 2

3 (8.314) S S T k m
H K M M = -------------------
2

3 (8.314) S S T k m
H (K M M) = -------------------
2

3 (8.314) S S T k m
H = -------------------
2 K M M

H = 18,270.015

18,270.015 is THE HEAT ENERGY IN [ J ].

You enter...
Find the heat energy measured in Joules given that the mass is 7.05[gm], deltaT is 75, and the specific heat is the specific heat of iron.
AutoMathic understands...
   Let 'H' stand for "HEAT ENERGY"

Let 'K' stand for "KILOGRAM"

Let 'M' stand for "METERS"

Let 'S' stand for "SECOND"

(Find HKMM/(SS))

Let 'm' stand for "MASS"

Let 'G' stand for "GRAM"

So...
7.05
m = ----
G


Let 'D' stand for "DELTAT"

D = 75

D = 75

Let 's' stand for "SPECIFIC HEAT"

So...
450 S S
s = -------
M M


Anything else?

> No.

I know that THERE ARE A THOUSAND GRAMS IN A KILOGRAM
I know that MOLARITY IS MOLES PER LITER
I know that MOLES IS GRAMS / FORMULA WEIGHT
I know that PRESSURE IN [ATM] X VOLUME IN [L] = MOLES X 0.0821 X TEMPERATURE IN [KELVIN]
I know that HEAT ENERGY IN [JOULES] IS 3/2 X MOLES X 8.314 X TEMPERATURE IN [KELVIN]. ASSUME SECONDS=1; METERS=1
I know that HEAT ENERGY IS SPECIFIC HEAT TIMES MASS TIMES DELTAT. ASSUME SECONDS=1; METERS=1

So...
G
1000 = -
K


Let 'O' stand for "MOLARITY"

Let 'o' stand for "MOLES"

Let 'L' stand for "LITER"

So...
o
O = -
L


Let 'F' stand for "FORMULA WEIGHT"

So...
G
o = -
F


Let 'P' stand for "PRESSURE"

Let 'A' stand for "ATMOSPHERE"

Let 'V' stand for "VOLUME"

Let 'T' stand for "TEMPERATURE"

Let 'k' stand for "KELVIN"

So...
A L P V = 0.0821 T k o


So...
H K M M 3 (8.314) T k o
------- = ---------------
S S 2


S = 1

S = 1

M = 1

M = 1

450 S S
s = -------
M M

s = 450

So...
H = D m s


S = 1

S = 1

Undefining 's' ...

Undefining 'M' ...

M = 1

M = 1

450 S S
s = -------
M M

s = 450

Hypothetically...

K = 1

Solving for G:

G
1000 = -
K

G
- = 1000
K

1
G - = 1000
K

G = 1000 K

G = 1,000

7.05
m = ----
G

m = 0.00705

H = D m s

H = 237.9375

237.9375 is THE HEAT ENERGY MEASURED IN JOULES.

You enter...
What's the heat capacity in [kJ] when the volume is 4[L], the density is the density of water, and the specific heat is the specific heat of water?
AutoMathic understands...
   Let 'H' stand for "HEAT CAPACITY"

Let 'K' stand for "KILOGRAM"

Let 'M' stand for "METERS"

Let 'S' stand for "SECOND"

(Find HKMM/(1000SS))

Let 'V' stand for "VOLUME"

Let 'L' stand for "LITER"

So...
4
V = -
L


Let 'D' stand for "DENSITY"

Let 'G' stand for "GRAM"

Let 'm' stand for "MILLILITER"

So...
m
D = -
G


Let 's' stand for "SPECIFIC HEAT"

So...
4180 S S
s = --------
M M


Anything else?

> No.

I know that THERE ARE A THOUSAND GRAMS IN A KILOGRAM
I know that THERE ARE A THOUSAND MILLILITERS IN A LITER
I know that MOLARITY IS MOLES PER LITER
I know that MOLES IS GRAMS / FORMULA WEIGHT
I know that DENSITY IS THE RATIO OF MASS TO VOLUME
I know that PRESSURE IN [ATM] X VOLUME IN [L] = MOLES X 0.0821 X TEMPERATURE IN [KELVIN]
I know that HEAT ENERGY IN [JOULES] IS 3/2 X MOLES X 8.314 X TEMPERATURE IN [KELVIN]. ASSUME SECONDS=1; METERS=1
I know that HEAT ENERGY IS SPECIFIC HEAT TIMES MASS TIMES DELTAT. ASSUME SECONDS=1; METERS=1
I know that HEAT CAPACITY IS MASS TIMES SPECIFIC HEAT

So...
G
1000 = -
K


So...
m
1000 = -
L


Let 'O' stand for "MOLARITY"

Let 'o' stand for "MOLES"

So...
o
O = -
L


Let 'F' stand for "FORMULA WEIGHT"

So...
G
o = -
F


Let 'A' stand for "MASS"

So...
A
D = -
V


Let 'P' stand for "PRESSURE"

Let 'a' stand for "ATMOSPHERE"

Let 'T' stand for "TEMPERATURE"

Let 'k' stand for "KELVIN"

So...
L P V a = 0.0821 T k o


Let 'h' stand for "HEAT ENERGY"

So...
K M M h 3 (8.314) T k o
------- = ---------------
S S 2


S = 1

S = 1

M = 1

M = 1

4180 S S
s = --------
M M

s = 4,180

Let 'd' stand for "DELTAT"

So...
h = A d s


S = 1

S = 1

Undefining 's' ...

Undefining 'M' ...

M = 1

M = 1

4180 S S
s = --------
M M

s = 4,180

So...
H = A s


Hypothetically...

m = 1

Solving for L:

m
1000 = -
L

m
- = 1000
L

m = 1000 L

1000 L = m

L (1000) = m

m
L = ----
1000

L = 0.001

4
V = -
L

V = 4,000

Hypothetically...

K = 1

Solving for G:

G
1000 = -
K

G
- = 1000
K

1
G - = 1000
K

G = 1000 K

G = 1,000

m
D = -
G

D = 0.001

Solving for A:

A
D = -
V

A
- = D
V

1
A - = D
V

A = D V

A = 4

H = A s

H = 16,720

16.72 (or 418 / 25, or 16 & 18 / 25) is THE HEAT CAPACITY IN [ KJ ].