To express quantitatively the different quantities relating to the study of physical phenomena, we use the units of measurement of the international system.
Some of these units are certainly familiar to you such as the meter or the kilogram ; others will probably be unknown to you because they are used here for the first time.
The best-known units are represented by their symbol (m = meter, kg = kilogram, s = second ...) ; the others are indicated with their symbol as well as with their name written in full in parentheses.
FORMULA 37
- Calculation of the speed of a body knowing the length and duration of the course (time).
Enunciated :
The speed expressed in meters per second is equal to the ratio between the distance traveled in meters and the duration of the journey expressed in seconds.
Enunciated
: The acceleration expressed in meters per second per second (or meters per square second) is equal to the ratio between the variation in speed expressed in meters per second and the time expressed in seconds.
If the variation is positive (increase in speed), we have a positive acceleration called simply acceleration ; if on the other hand, the variation is negative (reduction of the speed), one has a negative acceleration called deceleration.
Generally, values that express deceleration are preceded by the sign - (minus).
Examples :
a) positive acceleration :
Data : v" = 120 m /s ; v' =40 m /s ; t = 10 s
Speed variation : v"- v' = 120 -40=80 m /s
Acceleration : a = 80 / 10 = 8 m /s2
b) negative acceleration :
Data : v" = 30 m/ s ; v' = 90 m /s ; t = 5 s
Speed variation : v"- v' = 30 -90 = -60 m / s
Deceleration : a = - (60/5) = - 12 m/ s2
FORMULA 41
- Calculation of the acceleration of a body having a given mass and subjected to the action of a given constant force.
Enunciated :
The acceleration expressed in meters per second per second (or meters per square second) is equal to the ratio between the force expressed in newtons and the mass expressed in kilograms.
a = F/ m
a = acceleration in m / s2 (meters per second per second)
F =force in N (newtons)
m = mass in kg
Example :
Data : F =20 N ; m =0,5 kg
Acceleration : a = 20 / 0,5= 40 m/ s2
FORMULA 42
- Calculation of the constant force acting on a body having a given mass and a given acceleration.
Enunciated
: The weight, expressed in newtons, is equal to the product of mass expressed in kilograms by the value of gravity. This is worth about 9.81 m / s2
P
9,81 . m
P = weight in N (newtons)
m = mass in kg
Example :
Data : m = 15 kg
Weight : P
9,81 x 15 =147,15 N
FORMULA 45 -
Calculation of the specific mass of a body having a given mass and a given volume.
The term "specific weight" is used more correctly and sometimes the denomination "absolute density" instead of that of specific mass.
Enunciated :
The specific mass of a body, expressed in kilograms per cubic meter, is the ratio of mass in kilograms to volume in cubic meters.
ms = m /v
ms =specific gravity in kg / m3 (kilogram per cubic meter)
m = mass in kg
v = volume in m3
Example :
Data : m = 1998,26 kg d'eau ; v = 2 m3
Specific mass of water : ms =1998,26 /2 = 999,13 kg / m3
OBSERVATION:
Generally, the specific mass of bodies is expressed in grams per cubic centimeter (g / cm3) or in kilograms per cubic decimetre (kg / dm3).
1 g /cm3 = 1 kg /dm3 =1 000 kg /m3
1 kg/m3 =0,001 kg / dm3 = 0,001 g /cm3
In Table I (Figure 8), the specific masses of some bodies are indicated. All values are in kilograms per cubic decimeter ; the values for solids and liquids, excluding gases and mercury, are those corresponding to the ambient temperature of 15° C ; those of gases and mercury are those which correspond to the temperature of 0° C.
FORMULA 46
- Calculation of the volume of a body knowing its mass and its specific mass.
ms = density in kg/ dm3 (kilogram per cubic decimetre)
v = volume in dm3
(This formula is from formula 45).
Example :
Data : specific mass of iron, ms = 7,86 kg/dm3 ; volume of an iron cube, v = 0,02 dm3
Mass of the cube : m = 7,86 x 0,02 =0,1572 kg
FORMULA 48
- Calculation of the specific heat of a body knowing its mass and the amount of heat that must be supplied to this body to obtain a determined temperature increase.
Enunciated
: The specific heat, expressed in kilocalories per kilogram per degree Celsius, is equal to the ratio between the amount of heat expressed in kilocalories and the product of the mass expressed in kilograms by the increase in temperature expressed in degrees Celsius.
In Table II (Figure 9) the specific heats of some bodies are indicated. All values are in kilocalories per kilogram per degree Celsius (kcal / kg° C) and are considered to be essentially constant for temperatures between 0° C and 100° C unless otherwise indicated.
FORMULA 49
- Calculation of the amount of heat that must be supplied to a specific body of heat and mass known to achieve a given temperature increase.
Qc = quantity of heat transferred in kcal (kilocalories)
Cs= specific heat in kcal/kg . °C (kilocalories per kilogram per degree Celsius)
m = mass in kg
t"= final temperature in °C
t' =initial temperature in °C
t" - t' = temperature increase in °C
(This formula is from formula 48).
Example :
Data : specific heat of copper Cs = 0,093 kcal /kg . °C ; m = 25 kg of copper ;
t" =60 °C ; t' = 40 °C
Temperature increase : t" -t' = 60 - 40 =20 °C
Amount of heat needed : Qc = 0,093 x 25 x 20 = 46,50 kcal
FORMULA 50
- Calculation of the temperature increase of a body to which a given quantity of heat is transferred knowing the mass and specific heat of the body (in the calculation, we do not take into consideration the possible heat losses that may occur during the heat input).
Value in degrees Fahrenheit : tf = (1,8 x 20) + 32 = 36 + 32 = 68 °F
FORMULA 52
- Calculation of the value of a temperature in degrees Celsius knowing the value of this temperature in degrees Fahrenheit (unit of measurement of the English system).
Value in degrees Celsius : tc = 5 / 9 x (68 - 32) = 5 / 9 x 36 = 180 / 9 = 20 °C
FORMULA 53
- Calculation of the value of a temperature in kelvin (unit of measurement of the absolute temperature) knowing the value of this temperature in degrees Celsius.
Enunciated
: The mechanical work, expressed in Joules, is equal to the product of the force expressed in newtons by the length of the displacement of the force expressed in meters.
W = F . l
W = mechanical work in J (joules)
F = force en N (newtons)
l = length of movement in m (meters)
Example :
Data : F = 5 N ; l = 6 m
Mechanical work : W = 5 x 6 = 30 J
FORMULA 56
- Calculation of the kinetic energy of a moving body knowing its mass and speed at the moment considered.
Enunciated
: The kinetic energy, expressed in Joules, is equal to the half-product of the mass expressed in kilograms by the square of the speed expressed in meters per second.
Ec = (1 / 2) . m . v2
Ec = kinetic energy in J (joules)
m = mass in kg
v = speed in m / s
Example:
Data : m = 0,25 kg ; v = 7,5 m / s
Kinetic energy : Ec = (1 / 2) x 0,25 x 7,52 = 7,03 J
FORMULA 57
- Calculation of the consumed power knowing the quantity of energy absorbed during a given duration.
Enunciated :
The amount of heat (expressed in kilocalories) obtained by the total thermal transformation of mechanical work approximately equal to the product of this work (expressed in joules) by the number 0.000238.
Qc = 0,000238 . W
Qc = amount of heat (heat energy) in kcal (kilocalories)
W = mechanical work (mechanical energy) in J (joules)
Example :
Data : W = 625 000 J (joules)
Amount of heat obtained : Qc
0,000238 x 625 000 = 148,75 kcal
OBSERVATION :
The number 0.000238 is not a fixed coefficient.
Indeed, it indicates how many kilocalories can be obtained from a mechanical work (that is to say the corresponding energy) expressed in joules. This number therefore depends on the unit of measure chosen for the mechanical energy ; using the calorie instead of the kilocalorie, one must replace the number 0.000238 by the number 0.238.
FORMULA 60
- Calculation of the mechanical work (mechanical energy) corresponding to a given quantity of heat (thermal energy).
Enunciated
: The mechanical work (expressed in joules) corresponding to a given amount of heat (expressed in kilocalories) is approximately equal to the product of the amount of heat by the number 4 200.
Speed of a body
9,81 . m
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