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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.
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
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
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
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).
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 / s^{2}
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 / s^{2}
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 / s^{2} (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 / s^{2}
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 / s^{2} (meters per second per second)
Example :
Data : m = 0,5 kg ; a = 40 m / s^{2}
Force : F = 0,5 x 40 = 20 N
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 / s^{2} (meters per second per second)
(This formula is from formula 41).
Example :
Data : F = 20 N ; a = 40 m / s^{2}
Mass : m = 20 / 40 = 0,5 kg
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 / s^{2}
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.
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.
m_{s} = m / v
m_{s} = specific gravity in kg / m^{3} (kilogram per cubic meter)
m = mass in kg
v = volume in m^{3}
Example :
Data : m = 1998,26 kg d'eau ; v = 2 m^{3}
Specific mass of water : m_{s} = 1998,26 / 2 = 999,13 kg / m^{3}
OBSERVATION :
Generally, the specific mass of bodies is expressed in grams per cubic centimeter (g / cm^{3}) or in kilograms per cubic decimetre (kg / dm^{3}).
1 g / cm^{3} = 1 kg / dm^{3 }= 1 000 kg / m^{3}
1 kg / m^{3 }= 0,001 kg / dm^{3} = 0,001 g /cm^{3}
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.
Calculate the volume of a body
v = m / m_{s}
v = volume in dm^{3}
m = mass in kg
m_{s }= specific gravity in kg / m^{3} (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, m_{s} = 7,86 kg / dm^{3}
Chassis volume : v = 0,175 / 7,86 0,022265 dm^{3} = 22,265 cm^{3 }= 22 265 mm^{3}
FORMULA 47 - Calculation of the mass of a body knowing its specific mass and its volume.
m = m_{s} . v
m = mass in kg
m_{s} = density in kg / dm^{3} (kilogram per cubic decimetre)
v = volume in dm^{3}
(This formula is from formula 45).
Example :
Data : specific mass of iron, m_{s} = 7,86 kg / dm^{3} ; volume of an iron cube, v = 0,02 dm^{3}
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.
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.
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.
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
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
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
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
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
FORMULA 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)
Example :
Data : Tk = 293,16 K
Value in degrees Celsius : tc = 293,16 - 273,16 = 20 °C
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
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 . v^{2}
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 7,5^{2} = 7,03 J
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
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
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.
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
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