Formulas 37 to 60   “Physical”
Speed of a body Distance covered by a body Time put by a body
Calculation of the acceleration of a body The acceleration of a body having a mass  Force constant which acts on a body 
Mass of a body moving Weight of a body having a mass  Specific mass of a body 
Volume of a body  Calculation of the mass of a body  Specific heat of a body 
Calculation of the quantity of heat  Calculation of the increase in temperature  Temperature in degree Fahrenheit 
Temperature in degree Celsius (1) Temperature in Kelvin   Temperature in degree Celsius (2) 
Mechanical work relating to a body   Kinetic energy of a body  Calculation of the consumption 
Calculation of absorptive energy Calculation of the quantity of heat  Calculation of mechanical work
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Created it, 05/10/15

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Reception

MATHEMATICAL FORMS   “PHYSICAL”   2nd PART

PHYSICS 

Quantitatively to express the various sizes relating to the study of the physical phenomena, one has recourse to the measuring units of the international system.

Some of these units are certainly familiar for you such as the meter or the kilogram; others probably will be unknown for you because they are used here for the first time.

The most known units are represented by their symbol (m = meter, kg = kilogram, S = second…) ; the others are indicated with their symbol like with their name written in all letters between brackets.

HIGH OF PAGE FORMULATE 37 - Calculation the speed of a body knowing the length and the duration of the course (time).

Statement : The speed expressed in meters a second is equal to the relationship between the distance covered expressed in meters and the duration of the course expressed in seconds.

v = d / t

v = speed in m / s

t = distance in m

t = time in s

Example :

Data : d = 64 m ; t = 16 s

Speed: v = 64 / 16 = 4 m / s

HIGH OF PAGE FORMULATE 38 - Calculation of the distance covered by a body knowing the speed and the duration of the course.

d = v . t

d = distance in m

v = speed in m / s

t = time in s

(This formula is drawn from formula 37).

Data : v = 4 / s ; t = 16 s

Outdistance : d = 4 x 16 = 64 m

HIGH OF PAGE FORMULATE 39 - Calculation of the time put by a body to traverse a distance given with a given speed.

t = d / v

t = time in s

d = distance in m

v = speed in m / s

(This formula is drawn from formula 37).

Data : d = 64 m ; v = 4 m / s

Time : t = 64 / 4 = 16 s

HIGH OF PAGE FORMULATE 40 - Calculation of the acceleration of a body knowing the variation speed and the time during which this variation is carried out.

Statement : The acceleration expressed in meters a second a second (or meters a second square) is equal to the relationship between the variation the speed expressed in meters a second and the time expressed in seconds.

If the variation is positive (increase in speed), there is a positive acceleration simply called acceleration; so on the other hand, the variation is negative (reduction speed), one has a negative acceleration called deceleration.

Generally, the values which express a deceleration are preceded by the sign - (less).

 formule40

Examples :

a) positive acceleration:

Data : v" = 120 m / s ; v' = 40 m / s ; t = 10 S

Variation speed : 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

Variation speed: v" - v' = 30 - 90 = - 60 m / s

Deceleration: a = - (60 / 5) = - 12 m / s2

HIGH OF PAGE FORMULATE 41 - Calculation of the acceleration of a body having a mass given and subjected to the action of a given constant force.

Statement : The acceleration expressed in meters a second a second (or meters a second square) is equal to the relationship between the force expressed in newtons and the mass expressed in kilograms.

a = F / m

a = acceleration in m / s2 (meters a second a second)

F = force in N (newtons)

m = mass in kg

Data : F = 20 N ; m = 0,5 kg

Acceleration: a = 20 / 0,5 = 40 m / s2

HIGH OF PAGE FORMULATE 42 - Calculation of the constant force which acts on a body having a given mass and a given acceleration.

F = m . a

F = force in N (newtons)

m = mass in kg

a = acceleration in m / s2 (meters a second a second)

Data : m = 0,5 kg ; a = 40 m / s2

Force : F = 0,5 x 40 = 20 N

HIGH OF PAGE FORMULATE 43 - Calculation of the mass of a body moving subjected to the action of a given force and having a given acceleration.

m = F / a

m = mass in kg

F = force in N (newtons)

a = acceleration in m / s2 (meters a second a second)

(This formula is drawn from formula 41).

Data : F = 20 N ; a = 40 m / s2

Mass : m = 20 / 40 = 0,5 kg

HIGH OF PAGE FORMULATE 44 - Calculation of the weight of a body having a given mass and being subjected to the action of terrestrial gravity

Statement : The weight, expressed in newtons, is equal to the product bulk expressed in kilograms by the value of terrestrial gravity. This one is worth approximately 9,81 m / s2

P 9,81 . m

P = weight in N (newtons)

m = mass in kg

Data : m = 15 kg

Weight : P 9,81 x 15 = 147,15 N

HIGH OF PAGE FORMULATE 45 - Calculation of the specific mass of a body having a given mass and a given volume.

One more correctly uses the denomination “weight unit” and sometimes the denomination “density absolute” instead of that of specific mass.

Statement : The specific mass of a body, expressed in kilograms per cubic meter, is equal to the relationship between the mass expressed in kilograms and the volume expressed in cubic meters.

ms = m / v

ms = specific mass in kg / m3 (kilogram per cubic meter)

m = mass in kg

v = volume in m3

Data : m = 1998,26 kg of water ; v = 2 m3

Specific mass of water: ms = 1998,26 / 2 = 999,13 kg / m3

OBSERVATION :

Generally, the specific mass of the 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 the values are expressed in kilograms per cubic decimetre; the values relating to the solids and the liquids, other than gases and of mercury, are those corresponding to the ambient temperature of 15 °C; those of gases and mercury are on the other hand those which correspond to the temperature of 0 °C.

HIGH OF PAGE FORMULATE 46 - Calculation of the volume of a body knowing its mass and its specific mass.

v = m / ms

v = volume in dm3

m = mass in kg

ms = specific mass in kg / m3 (kilogram per cubic meter)

(This formula is drawn from formula 45).

Data : mass of a frame out of iron, m = 0,175 kg ; specific mass of iron, ms = 7,86 kg / dm3

Volume of the frame : v = 0,175 / 7,86 0,022265 dm3 = 22,265 cm3 = 22 265 mm3

Masse_specifique

HIGH OF PAGE FORMULATE 47 - Calculation of the mass of a body knowing its specific mass and its volume.

m = ms . v

m = mass in kg

ms = specific mass in kg / dm3 (kilogram per decimetre cubes)

v = volume in dm3

(This formula is drawn from formula 45).

Data : specific mass of iron, ms = 7,86 kg / dm3 ; volume of an iron cube, v = 0,02 dm3

Mass cube: m = 7,86 x 0,02 = 0,1572 kg

HIGH OF PAGE FORMULATE 48 - Calculation of the specific heat of a body knowing its mass and the quantity of heat which it is necessary to provide to this body to obtain an increase in given temperature.

Statement : The specific heat, expressed in kilogram calories per kilogram per degree Celsius, is equal to the relationship between the quantity of heat expressed in kilogram calories and the product bulk expressed in kilograms by the increase in temperature expressed in degree Celsius.

formule48

In table II (figure 9) the specific heats of some bodies are indicated. All the values are expressed in kilogram calories per kilogram per degree Celsius (kcal / kg . °C) and is practically regarded as constants for temperatures ranging between 0 °C and 100 °C unless otherwise specified.

Valeur_moyenne_de_la_chalor

HIGH OF PAGE FORMULATE 49 - Calculation of the quantity of heat which must be provided to a body of specific heat and mass known to obtain an increase given in temperature.

Qc = Cs . m . (T" - t')

Qc = quantity of heat yielded in kcal (kilogram calories)

Cs = specific heat in kcal / kg . °C (kilogram calories per kilogram per degree Celsius)

m = mass in kg

t" = final temperature in °C

t' = initial temperature in °C

t" - t' = increase in temperature in °C

(This formula is drawn from formula 48).

Data : specific heat of copper Cs = 0,093 kcal / kg . °C ; m = 0,25 kg of copper ;

t" = 60 °C ; t' = 40 °C

Increase in temperature : t" - t' = 60 - 40 = 20 °C

Quantity of heat necessary : Qc = 0,093 x 25 x 20 = 46,50 kcal

HIGH OF PAGE FORMULATE 50 - Calculation of the increase in temperature of a body to which one yields a given quantity of heat knowing the mass and the specific heat of the body (in calculation, one does not take into account the possible losses of heat which can occur at the time of the contribution of heat).

t" - t' = Qc / (m . Cs)

t" = final temperature in °C

t' = initial temperature in °C

t" - t' = increase in temperature in °C

Qc = quantity of heat yielded in kcal (kilogram calories)

m = mass in kg

Cs = specific heat in kcal / kg . °C (kilogram calories per kilogram per degree Celsius)

 

Data : Qc = 2 kcal ; m = 4 kg of iron ; specific heat of iron Cs = 0,118 kcal / kg . °C

Increase in temperature : t" - t' = 2 / (4 x 0,118) = 2 / 0,472 4,23 °C

If the initial temperature is 10 °C, the final temperature will be of 14,24 °C

HIGH OF PAGE FORMULATE 51 - Calculation of the value of a temperature in degree Fahrenheit (measuring unit of the English system) knowing the value of this temperature in degree Celsius.

tf = (1,8 x tc) + 32

tf = temperature in °F (degree Fahrenheit)

tc = temperature in °C (degree Celsius)

 

Data : tc = 20 °C

Value in degree Fahrenheit : tf = (1,8 x 20) + 32 = 36 + 32 = 68 °F

HIGH OF PAGE FORMULATE 52 - Calculation of the value of a temperature in degree Celsius knowing the value of this temperature in degree Fahrenheit (measuring unit of the English system).

tc = 5 / 9 x (tf - 32)

tc = temperature in °C (degree Celsius)

tf = temperature in °F (degree Fahrenheit)

 

Data: tf = 68 °F

Value in degree Celsius : tc = 5 / 9 x (68 - 32) = 5 / 9 x 36 = 180 / 9 = 20 °C

HIGH OF PAGE FORMULATE 53 - Calculation of the value of a temperature in Kelvin (measuring unit of the absolute temperature) knowing the value of this temperature in degree Celsius.

Tk tc + 273,16

Tk = temperature in K (Kelvin)

tc = temperature in °C (degree Celsius)

 

Data : tc = 20 °C

Value in Kelvin: Tk 20 + 273,16 = 293,16 K

HIGH OF PAGE FORMULATE 54 - Calculation of the value of a temperature in degree Celsius knowing the value of this temperature in Kelvin.

tc Tk - 273,16

tc = temperature in °C (degree Celsius)

Tk = temperature in K (Kelvin)

 

Data : Tk = 293,16 K

Value in degree Celsius : tc = 293,16 - 273,16 = 20 °C

HIGH OF PAGE FORMULATE 55 - Calculation of mechanical work relating to a body which moves under the action of a force directed in the direction of displacement.

Statement : 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. I

W = work mechanical in J (joules)

F = force in N (newtons)

I = length of displacement in m (meters)

 

Data: F = 5 N ; I = 6 m

Mechanical work: W = 5 x 6 = 30 J

HIGH OF PAGE FORMULATE 56 - Calculation of the kinetic energy of a body moving knowing its mass and its speed at the moment considered.

Statement : The kinetic energy, expressed in joules, is equal to the semi-finished product of the mass expressed in kilograms by the square the speed expressed in meters a second.

Ec = (1 / 2) . m . v2

Kinetic Ec = energy in J (joules)

m = mass in kg

v = speed in m / s

 

Data : m = 0,25 kg ; v = 7,5 m / s

Kinetic energy : Ec = (1 / 2) x 0,25 x 7,52 = 7,03 J

HIGH OF PAGE FORMULATE 57 - Calculation of the consumption knowing the quantity of energy absorptive for a given length of time.

Statement: The power, expressed in Watts, is equal to the relationship between the quantity of absorptive energy, expressed in joules, and the time expressed in seconds.

P = W / t

P = consumption out of W (Watts)

W = energy absorptive in J (joules)

t = time in s (seconds)

 

Data : W = 150 J ; t = 120 s

Power : P = 150 / 120 = 1,25 W

HIGH OF PAGE FORMULATE 58 - Calculation of absorptive energy knowing the consumption and the duration of consumption.

W = P. t

W = energy absorptive in J (joules)

P = consumption out of W (Watts)

t = lasted of consumption in s (seconds)

 

(This formula is drawn from formula 57).

Data : P = 1 500 W ; t = 3 600 s

Absorptive energy : W = 1 500 x 3 600 = 5 400 000 J

HIGH OF PAGE FORMULATE 59 - Calculation of the quantity of heat (thermal energy) correspondent with a given mechanical work (mechanical energy).

Statement : Quantity of heat (expressed in kilogram calories) obtained by the total thermal transformation of mechanical work roughly equal to the product of this work (expressed in joules) by number 0,000238.

Qc = 0,000238 . W

Qc = quantity of heat (thermal energy) in kcal (kilogram calories)

W = work mechanical (mechanical energy) in J (joules)

 

Data : W = 625 000 J (joules)

Quantity of heat obtained : Qc 0,000238 x 625 000 = 148,75 kcal

OBSERVATION :

Number 0,000238 is not a fixed coefficient.

Indeed, it indicates how much kilogram calories one can obtain from a mechanical work (i.e. of corresponding energy) expressed in joules. This number thus depends on the measuring unit chosen for the mechanical energy; by using the calorie instead of the kilogram calorie, one must replace number 0,000238 by number 0,238.

HIGH OF PAGE FORMULATE 60 - Calculation of mechanical work (mechanical energy) correspondent with a quantity of heat given (thermal energy).

Statement : Mechanical work (expressed in joules) correspondent with a quantity of heat given (expressed in kilogram calories) is roughly equal to the product of the quantity of heat by number 4 200.

W 4 200 . Qc

W = work mechanical in J (joules)

Qc = quantity of heat in kcal (kilogram calories)

 

Data : Qc = 30 kcal

Work corresponding : W = 4 200 x 30 = 126 000 J

 

     

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