FRANCAIS francophone2.gif ANGLAIS

 

 

Created the, 12/06/2019

 Updated the, 02/01/2020

Visiteurs N°  


Home
Back to Main Sites New Blog Novelty Search engine Your Shopping Cart For Shopping Your Member Area Bookmarks, Your Favorite Games Static Welcome Page Site in French Web Site in English
Summaries
Basic Electronics Fundamental Technology Test your Knowledge Digital Theoretical Electronics Digital Practical Electronics Digital Electronic Lexicon Data book TTL Data book CMOS TVC Troubleshooting Mathematical
Microcomputers
Theoretical of Microcomputers Test your Knowledge Practical Microcomputers Computer Glossaries
Physical
The light Field of Action Electromagnetic Radiation
Technologies
Classification of Resistances Identification of Resistances Classification of Capacitors Identification of Capacitors
Mathematical Forms
Geometry Physical 1. - Electronic 1. 2. - Electronic 1. 3. - Electrical 1. 4. - Electromagnetism
Access to all our Products
E. T. F. - Volume I - 257 Pages E. T. F. - Volume II - 451 Pages E. T. F. - Volume III - 611 Pages E. T. D. - Volume I - 610 Pages N. B. M. - Volume I - 201 Pages E. T. M. - Volume I - 554 Pages Business at Home Books 34 free pages Our E-books Geometry Software Electronic Components Software
Overview of all our Products
E. T. F. - Volume I - 257 Pages E. T. F. - Volume II - 451 Pages E. T. F. - Volume III - 611 Pages E. T. D. - Volume I - 610 Pages E. T. M. - Volume I - 554 Pages Geometry Software Electronic Components Software
Our Lessons in PDF Formats
Basic Electronics Fundamental Technology Digital Theoretical Electronics Digital Practical Electronics Theoretical of Microcomputers Mathematics
Data Processing
Troubleshooting Win98 and WinXP PC Troubleshooting Glossary HTML and Programs PHP and Programs JavaScript (in progress) Creation of several Sites
Forums
Electronic Forum and Infos Electronic Forum and Poetry
Miscellaneous and others
Form of the personal pages News XML Statistics CountUs JavaScript Editor Our Partners and Useful Links Partnership Manager Our MyCircle Partners Surveys 1st Guestbook 2nd Guestbook Site Directories




Bookmarks - Formulas 37 to 60 "Physics"   :
  Speed of a body   Distance traveled by a body   Time put by a body
  Calculation of the acceleration of a body   Acceleration body having mass     Force acting on a body
  Mass of a moving body   Weight of a body having mass   Specific mass of a body 
  Volume of a body   Calculating the mass of a body   Specific heat of a body
  Calculation of the amount of heat   The increase in temperature   Temperature in degrees Fahrenheit
  Temperature in degree Celsius (1)   Kelvin temperature   Temperature in degree Celsius (2)
  Mechanical work relating to a body   The kinetic energy of a body   The power consumed
  Calculation of absorbed energy   Calculation of the amount of heat   Calculation of mechanical work
   Footer  


Form of Physics :

PHYSICAL 

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.

HAUT DE PAGE 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.

Calculate the speed expressed in meters per second

v = d / t

v = speed in m / s

d = distance in m

t = time in s

Example :

Data : d = 64 m ; t = 16 s

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

HAUT DE PAGE FORMULA 38 - Calculation of the distance traveled by a body knowing the speed and duration of the course.

Calculate the distance traveled by a body

d = v . t

d = distance en m

v = speed in m / s

t = time in s

(This formula is taken from Form 37).

Example :

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

Distance : d = 4 x 16 = 64 m

HAUT DE PAGE FORMULA 39 - Calculation of the time taken by a body to travel a given distance with a given speed.

Calculate the time taken by a body

t = d / v

t = time in s

d = distance en m

v = speed in m / s

(This formula is from Formula 37).

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

Time : t = 64 / 4 = 16 s

HAUT DE PAGE FORMULA 40 - Calculation of the acceleration of a body knowing the variation of speed and the time during which this variation takes place.

Calculate the acceleration or deceleration of a body, in meters per square second

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).

 formule40

  • 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

HAUT DE PAGE FORMULA 41 - Calculation of the acceleration of a body having a given mass and subjected to the action of a given constant force.

Calculate the acceleration of a body, in meters per square second

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

HAUT DE PAGE FORMULA 42 - Calculation of the constant force acting on a body having a given mass and a given acceleration.

Calculate the constant force acting on a body

F = m . a

F = force in N (newtons)

m = mass in kg

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





  • Example :

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

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

HAUT DE PAGE FORMULA 43 - Calculation of the mass of a moving body subjected to the action of a given force and having a given acceleration.

Calculate the mass of a moving body

m = F / a

m = mass in kg

F = force in N (newtons)

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

(This formula is from formula 41).

  • Example :

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

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

HAUT DE PAGE FORMULA 44 - Calculation of the weight of a body having a given mass and being subjected to the action of gravity

Calculate the weight of a body with a given mass

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

HAUT DE PAGE FORMULA 45 - Calculation of the specific mass of a body having a given mass and a given volume.

Calculate the specific mass of a body

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.

HAUT DE PAGE FORMULA 46 - Calculation of the volume of a body knowing its mass and its specific mass.

Calculate the volume of a body

v = m / ms

v = volume in dm3

m = mass in kg

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

(This formula is from formula 45).

  • Example :

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

Chassis volume : v = 0,175 / 7,86 0,022265 dm3 = 22,265 cm3 = 22 265 mm3

Masse_specifique

HAUT DE PAGE FORMULA 47 - Calculation of the mass of a body knowing its specific mass and its volume.

Calculate the mass of a body

m = ms . v

m = mass in kg

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

HAUT DE PAGE 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.

Calculate the specific heat of a body

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.

formule48

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.

Valeur_moyenne_de_la_chaleur

HAUT DE PAGE 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.

Calculate how much heat a body needs

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

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

HAUT DE PAGE 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).

Calculate the temperature increase of a body

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

t" = final temperature in °C

t' = initial temperature in °C

t" - t' = temperature increase in °C

Qc = quantity of heat transferred in kcal (kilocalories)

m = mass in kg

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

  • Example :

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

Temperature increase : 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 14.24° C

HAUT DE PAGE FORMULA 51 - Calculation of the value of a temperature in degrees Fahrenheit (knowing the value of this temperature in degree Celsius).

Calculate the value of a temperature in degrees Fahrenheit

tf = (1,8 x tc) + 32

tf = temperature in °F (degree Fahrenheit)

tc = temperature in °C (degree Celsius)

  • Example :

Data : tc = 20 °C

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

HAUT DE PAGE 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).

Calculate the value of a temperature in degrees Celsius

tc = 5 / 9 x (tf - 32)

tc = temperature in °C (degree Celsius)

tf = temperature in °F (degree Fahrenheit)

  • Example :

Data : tf = 68 °F

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

HAUT DE PAGE 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.

Calculate the value of a temperature in Kelvin

Tk tc + 273,16

Tk = temperature in K (kelvin)

tc = temperature in °C (degree Celsius)

  • Example :

Data : tc = 20 °C

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

HAUT DE PAGE FORMULA 54 - Calculation of the value of a temperature in degree Celsius knowing the value of this temperature in kelvin.

Calculate the value of a temperature in degrees Celsius knowing the value of this temperature in Kelvin

tc Tk - 273,16

tc = temperature in °C (degree Celsius)

Tk = temperature in K (kelvin)

  • Example :

Data : Tk = 293,16 K

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

HAUT DE PAGE FORMULA 55 - Calculation of the mechanical work relative to a body which moves under the action of a force oriented in the direction of displacement.

Calculate the value of the mechanical work relating to a body

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

HAUT DE PAGE FORMULA 56 - Calculation of the kinetic energy of a moving body knowing its mass and speed at the moment considered.

Calculate the kinetic energy of a moving body

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

HAUT DE PAGE FORMULA 57 - Calculation of the consumed power knowing the quantity of energy absorbed during a given duration.

Calculate the power consumed knowing the amount of energy

Enunciated : The Power, expressed in watts, is the ratio of the amount of energy absorbed, expressed in joules, to the time in seconds.

P = W / t

P = power consumed in W (watts)

W = energy absorbed in J (joules)

t = time in s (seconds)

  • Example :

Data : W = 150 J ; t = 120 s

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

HAUT DE PAGE FORMULA 58 - Calculation of absorbed energy knowing the power consumed and the duration of consumption.

Calculate the energy absorbed knowing the power consumed and the duration of consumption

W = P . t

W = energy absorbed in J (joules)

P = power consumed in W (watts)

t = duration of consumption in s (seconds)

(This formula is taken from Form 57).

  • Example :

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

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

HAUT DE PAGE FORMULA 59 - Calculation of the amount of heat (thermal energy) corresponding to a given mechanical work (mechanical energy).

Calculate the amount of heat (thermal energy)

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.

HAUT DE PAGE FORMULA 60 - Calculation of the mechanical work (mechanical energy) corresponding to a given quantity of heat (thermal energy).

Calculate mechanical work (mechanical 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.

W 4 200 . Qc

W = mechanical work in J (joules)

Qc = amount of heat in kcal (kilocalories)

  • Example :

Donnée : Qc = 30 kcal

Corresponding work : W = 4 200 x 30 = 126 000 J









Nombre de pages vues, à partir de cette date : le 27 Décembre 2019
compteur visite blog gratuit


Mon audience Xiti



Send an email to Corporate Webmaster for any questions or comments about this Web Site.

Web Site Version : 11. 5. 12 - Web Site optimization 1280 x 1024 pixels - Faculty of Nanterre - Last modification : JANUARY 02, 2020.

This Web Site was Created on, 12 JUNE 2019 and has Remodeled, in JANUARY 2020.