import{E as Mt,c as xt,aj as vt,ag as Lt,r as $,b4 as a,w as t,f as V,b9 as at,ci as bt,a2 as Dt,U as C,m as ot,jL as It,jZ as Ct,l as Ft}from"./index-ajJ0B2-K.js";import{t as Ht}from"./deprecatedFractions-MjvQvhWQ.js";import{k as y}from"./message-pLsPPsNT.js";import{b as pt,p as lt}from"./Personne-zmVSe1SX.js";import{t as v}from"./style-YtQgMMZt.js";import"./modales-xge8ru8I.js";const St="Solve problems with compound quantities and complex unit conversions",Qt="72c5a",Bt="4P10";function zt(){Mt.call(this),this.titre=St,this.consigne="",this.nbQuestions=3,this.nbCols=1,this.nbColsCorr=1,xt.isHtml?this.spacing=3:this.spacing=1.5,xt.isHtml?this.spacingCorr=3:this.spacingCorr=2,this.sup="",this.nouvelleVersion=function(k){this.listeQuestions=[],this.listeCorrections=[];const Tt=vt([0,1,2],this.nbQuestions);let gt=0,kt=0;const qt=vt([0,1],this.nbQuestions);let ft,S,dt,N,z,M,L,A,D,G,I,Q,_,P,K,E,ct,rt,$t,nt,wt,m,tt,et,R,ht,U,j;const Wt=Lt({saisie:this.sup,max:14,melange:15,defaut:15,nbQuestions:this.nbQuestions});let w=1;xt.isHtml||(w=0);const F=[["salt","water",300],["sugar","water",2e3],["carbon dioxide","water",3],["sodium bicarbonate","water",9],["sodium carbonate","water",300]],W=[["palladium",12e3," of "," of "],["steel",7800," of the ",""],["melting",7100," of the "," of "],["aluminum",2700," of the ",""],["money",10500," of the ",""],["bronze",8800," of "," of "],["copper",8960," of "," of "],["iron",7860," of "," of "],["lithium",530," of "," of "],["mercury",13545,"of ","of "],["nickel",8900," of "," of "],["gold",19300," of the ",""],["platinum",21450," of "," of "],["titanium",4500," of "," of "],["zinc",7150," of "," of "]],b=[["Nice",342637,71.9],["Montpellier",281613,56.9],["Rennes",216268,50.4],["Dijon",155090,40.4],["Orleans",114782,27.5],["Clermont-Ferrand",142686,42.7],["Nantes",306694,65.2],["Paris",2190327,105.4],["Lyon",515695,47.9],["Marseille",862211,240.6],["Bordeaux",252040,49.4],["Nancy",104592,15],["Toulouse",475438,118.3],["Lille",232440,34.8],["Strasbourg",279284,78.3]],it=[["A bike",1.5,2,8],["a canoe",10,2,4],["roller-skate",7,2,5],["a sand yacht",12,2,4]],Z=[["piano",20],["mathematics",25],["yoga",5],["of drawing",12],["sailing",15]],u=[["peaches",4,10,30],["nut",5.4,4,13],["cherries",5.6,11,20],["apples",2.2,20,40],["raspberries",15,1,5],["strawberries",7.5,5,10],["lemons",1.5,15,30],["bananas",1.5,15,25]],H=[["radiator",2300,20],["TV",46,12],["electric oven",2960,4],["computer",460,8]],X=[["whole milk",1.032],["gasoline",.755],["diesel",.83],["oil",.91],["beer",.9],["sand",1.6]],T=[["Marl","Gournay-sur-Marne",110,550,"April 1983","there","of the"],["Seine","Alfortville",218,2100,"January 1982","there","of the"],["Oise","Pont-Sainte-Maxence",109,665,"February 1995","L'","of the'"],["Loire","Saint Nazaire",931,5350,"December 1999","there","of the"],["Rhine","Strasbourg",951,3310,"June 2016","THE"," of "],["Rhone","Beaucaire",1690,11500,"December 2003","THE"," of "],["Meuse","Chooz",144,1610,"January 1995","there","of the"]],J=[["on a bike",4,12,8],["In a train",50,100,5],["in a car",15,30,5],["by plane",150,250,12],["walk",2,4,5]];for(let p=0,st,f,s,i,r,l,c,x,g,h,n,d,O,mt,Y,B,ut,o,e,yt=0;p<this.nbQuestions&&yt<50;){switch(Wt[p]){case 1:f=$(0,3),ft=H[f][0],S=H[f][1],dt=H[f][2],N=$(0,3),c=$(dt/4,dt,[1]),r=c+N*.25,z=$(0,5)/100+.14,o=`The label on the back of a ${ft} indicates a power of $${t(S)}$ Watts. We run it for $${Math.floor(r)}$ hours`,N!==0&&(o+=`and $${N*15}$ minutes`),o+=`.<br>The price of one kWh is $${v(z)}$ €.<br>`,o+=a(0)+" Express in kWh"+y(k+p+1,w,"energy","Definition: Energy (physical quantity)","It is the product of electrical power (Watt) times time (s) and is measured in Joules (J).<br>1 J = 1 W × 1 s<br>However, to measure larger energies , we instead use the kiloWatt hour (kWh).<br>1 kWh = 1000 W × 1 h")+"consumed.<br>",o+=a(1)+" Calculate the corresponding expense.",e=a(0)+` A ${ft} with a power of $${t(S)}$ Watts that operates for $${Math.floor(r)}$ hours`,N!==0&&(e+=`and $${N*15}$ minutes`),e+=" consumes: <br>",N!==0&&(e+=`$${c}\\text{ h } ${N*15} = ${c}\\text{ h} + ${Ht(N,4)}\\text{ h} =${t(c+N*.25)}\\text{ h}$<br>`),e+=`$${S}\\text{ W}\\times${t(r)}\\text{ h}=${t(S/1e3)}\\text{ kW}\\times${t(r)}\\text{ h}=${t(S*r*.001,3)}\\text{ kWh}$<br>`,e+=a(1)+` The price of this energy consumed is: $${v(z)}$ €$\\text{/kWh} \\times${t(S*r*.001,3)}\\text{ kWh}`,z*S*r/10!==Math.round(z*S*r/10)?e+=`\\approx${v(z*S/1e3*r)}$ €.`:e+=`=${v(z*S/1e3*r)}$ €.`;break;case 2:switch(s=qt[kt],kt++,s){case 0:M=180+$(0,10)*10,L=80+$(0,4)*10,A=5+$(0,5),D=A*2+$(0,4)*2,G=$(2,5),o=`A swimming pool is shaped like a right prism. The depth at its north end is $${M}$ cm and the depth at its south end is $${L}$ cm.<br>`,o+=`From one end to the other, the slope at the bottom of the pool is regular.<br>The width of the pool (East-West) is $${A}$ m and its length (North-South) is $${D} $ m.<br>`,o+=a(0)+" Calculate the"+y(k+p*3,w,"volume","Definition: Volume (physical quantity)",`It is the product of three lengths or the product of an area and a length.<br>The unit of measurement for volume is the cubic meter (m${at(3)}) but we can also find the liter (L) with as correspondence 1dm${at(3)} = 1L.`)+` of water in m${at(3)} contained in this swimming pool when it is full.<br>`,o+=a(1)+` Knowing that to raise the temperature of a liter of water by 1 degree requires energy of $${t(1.162)}$ Watt-hour, what is the energy consumed in kWh to increase the temperature of the swimming pool by $${G}$ degrees?<br >`,e=a(0)+` The base of this right prism is a rectangular trapezoid of small base $${L}$ cm, of large base $${M}$ cm and of height $${D}$ m.<br>`,e+=`$\\mathcal{A}=\\dfrac{\\left(${M}\\text{ cm}+${L}\\text{ cm}\\right)}{2}\\times${D}\\text{ m}$`,e+=` $=\\dfrac{\\left(${t(M/100,2)}\\text{ m}+${t(L/100,2)}\\text{ m}\\right)}{2}\\times${D}\\text{ m}$`,e+=` $=\\dfrac{${t((M+L)/100,2)}\\text{ m}}{2}\\times${D}\\text{ m}$`,e+=` $=${t((M+L)/200,2)}\\text{ m}\\times${D}\\text{ m}$`,e+=` $=${t((M+L)/200*D,2)}\\text{ m}^2$<br>`,e+="The volume of this prism, and therefore, by extension, the volume of water contained in the swimming pool, is:<br>",e+=`$\\mathcal{A}\\times\\mathcal{h}=${t((M+L)/200*D,2)}\\text{ m}^2\\times${A}\\text{ m}$`,e+=` $=${t((M+L)/200*D*A,2)}\\text{ m}^3$.<br>`,e+=a(1)+` Let's convert the volume of the pool into liters: $${t((M+L)/200*D*A,2)}\\text{ m}^3=${t((M+L)*D*A*5)}\\text{ dm}^3=${t((M+L)*D*A*5)}\\text{ L}$<br>`,e+=` The energy consumed to raise the water temperature of this pool by $${G}$ degrees is:<br>`,e+=`$\\mathcal{E}=${t((M+L)*D*A*5)}\\text{ L}\\times${G}\\text{ °C}\\times 1.162 \\dfrac{\\text{Wh}}{\\text{°C}\\ times\\text{L}}=${t((M+L)*D*A*5*G*1.162,3)}\\text{ Wh}=${t((M+L)*D*A/200*G*1.162,7)}\\text{ kWh}$<br>`;break;case 1:i=$(0,5),I=$(10,15)*2,Q=$(0,10)+I*4,o=`A cylindrical barrel has a radius of $${I}$ cm and a height of $${Q}$ cm.<br>`,o+=a(0)+" Calculate the"+y(k+p*3,w,"volume","Definition: Volume (physical quantity)","It is the product of three lengths or the product of an area and a length.<br>The unit of measurement for volume is the cubic meter ($\\text{m}^3 $) but we can also meet the liter (L) with the correspondence $\\text{1dm}^3 = \\text{1L}.$")+` in dm${at(3)} to $${t(.1)}$ near this barrel.<br>`,o+=a(1)+` If we fill it in ${X[i][0]} (whose`+y(k+p*3,w,"density","Definition: Density (physical quantity)","The density of a substance is equal to the density of the substance divided by the density of the reference body at the same temperature.<br>For liquids and solids, water is used as a reference (its density is 1 kg/dm$^3$), for gases, the measurement is carried out relative to air.<br>So for liquids, the density is equal to the density expressed in kg/dm$^3$.")+` is $${t(X[i][1])}$ kg/dm$^3$), what mass will ${X[i][0]} in kg contain to the nearest gram?<br>`,e=a(0)+` The volume of a cylinder is given by the formula $\\mathcal{A}\\text{base ire}\\times\\mathcal{h}$.<br> Here, the base is a disk of radius $ ${I}$ cm.<br>`,e+=`$\\mathcal{A}\\text{base ire}\\times\\mathcal{h}=\\pi\\times${I}^{2}\\text{ cm}^2\\times${Q}\\text{ cm}=${t(I*I*Q)}\\pi\\text{ cm}^3\\approx${t(I*I*Q*Math.PI,1)}\\text{ cm}^3\\approx${t(I*I*Q*Math.PI/1e3,1)}\\text{ dm}^3$<br>`,e+=a(1)+` The ${X[i][0]} mass contained in this barrel is:<br>`,e+=`$${t(I*I*Q*Math.PI/1e3,1)}\\text{ dm}^3\\times ${t(X[i][1])} \\dfrac{kg}{dm^3}\\approx${t(I*I*Q*Math.PI/1e3*X[i][1],3)}\\text{ kg}.$`;break}break;case 3:l=lt(),s=$(0,4),d=$(40,70),h=$(J[s][1],J[s][2]),o=`${l} moves ${J[s][0]} to the`+y(k+p*3,w,"speed","Definition: Speed (physical quantity)","Speed is the quotient of the distance traveled by the travel time.<br>The official unit is meters per second ($\\text{m/s}$ or $\\text{m.s}^{- 1}$) but we often use kilometers per hour ($\\text{km/h}$ or $\\text{km.h}^{-1}$).")+` of $${t(h)}\\text{ m/s}$.<br>`,o+=`It weighs $${d}\\text{ kg}$.<br>`,o+=a(0)+" Calculate its"+y(k+p*3+1,w,"amount of movement","Definition: Quantity of movement (physical quantity)","It is the product of the mass of a body and its speed.<br>The unit of measurement for momentum is $\\text{kg.m.s}^{-1}$.")+" in $\\text{kg.m.s}^{-1}$.<br>",o+=a(1)+" Deduce its"+y(k+p*3+2,w,"kinetic energy","Definition: Kinetic energy (physical quantity)","The kinetic energy of a body of mass $m$ (in kg) assimilated to a material point moving at speed $v$ (in m/s) is given by the formula <br>$E=\\dfrac{1}{2}\\times m\\times v^2$.<br>The unit of measurement for kinetic energy is the Joule (J).<br>$1J = 1\\text{ kg.m}^2\\text{s}^{-2}$")+" in Joules.",e=a(0)+` The momentum of ${l} is: $${d} \\text{ kg}\\times ${h}\\text{ m/s}=${C(d*h)}\\text{ kg.m.s}^{-1}$.<br>`,e+=a(1)+` The kinetic energy of ${l} is: $\\dfrac{1}{2}\\times ${d} \\text{ kg}\\times (${h}\\text{ m/s})^2=\\dfrac{${d} \\times${h}^2}{2}\\text{ J}=${t(d*h**2/2)}\\text{ J}$.`;break;case 4:l=pt(),f=$(60,90),d=$(20,30),n=$(25,35)/10,o=`${l} who weighs $${d}$ kg is on the seat of a swing "`+y(k+p*3,2,"trebuchet","Explanatory diagram","images/trebuchet.png")+`" in a kindergarten. The seat is located $${t(n,1)}$ m from the central pivot of the swing (lever arm).<br>`,o+=a(0)+" Calculate the"+y(k+p*3+1,w,"moment","Definition: Moment (physical quantity)","The moment of a force of intensity $F$ (in Newton or kg.m.s$^{-2}$) at a point M with respect to a pivot P is the product of $F$ by the distance $ d=$PM (called lever arm) expressed in meters, or $F\\times d$ (when this force is exerted perpendicular to the lever arm). The moment is the energy allowing the object to rotate around the pivot.<br>The unit of measurement for the moment is the Joule (J).<br>$1J=1\\text{ kg. m}^2\\text{s}^{-2}$")+" of"+y(k+p*3+2,w,"weight","Definition: Weight (physical quantity)","Weight is the product of the mass $m$ of an object by the acceleration of the Earth's gravity ($g=9.81\\text{ m.s}^{-2}$).<br>L 	he unit of weight is the Newton (N): 1 N = 1 kg.m.s$^{-2}$")+` of ${l} on its seat in relation to the central pivot of the trebuchet in Joules (we will assume that the lever arm is horizontal).<br>`,o+=a(1)+` ${l}'s father comes to settle on the other side of the central pivot. It weighs $${f}$ kg and is installed so that its weight allows the swing to be balanced horizontally. How long should the lever arm be on its side? In other words, how far from the pivot is ${l}'s father sitting?<br>`,e=a(0)+` The moment of the weight of ${l} applied to its seat in relation to the central pivot of the trebuchet is:<br>`,s=d*9.81*n,e+=`$${d}\\text{ kg} \\times 9.81 \\text{m.s}^{-2} \\times ${t(n,1)} \\text{ m} = ${t(s,3)}\\text{ kg.m}^2\\text {.s}^{-2}=${t(s)}\\text{ J}$.<br>`,e+=a(1)+` In order to balance the trebuchet, the father of ${l} must position himself so that the moment of his weight on his seat point in relation to the central pivot of the trebuchet is equal to that of ${l}, we obtain the following equation where $ ${ot("d","black")}$ represents its distance from the central pivot:<br>`,e+=`$ ${f}\\text{ kg}\\times 9.81 \\text{m.s}^{-2} \\times ${ot("d","black")} \\text{ m}=${t(s)}\\text{ J}$.<br>`,e+=`Hence, $${ot("d","black")}\\text{ m} = \\dfrac{${t(s,3)}\\text{ J}}{${f}\\text{ kg}\\times 9.81 \\text{m.s}^{-2} }\\approx${t(s/(9.81*f),2)}\\text{ m}.$`;break;case 5:K=$(3,6),E=$(3,6,[K]),ct=$(5,8),rt=ct*E,$t=ct*K,o=a(0)+` A city bus carries on average $${rt}$ people at a time.<br> The average trip length is $${K}$ km.<br> Calculate the`+y(k+p*3,w,"traffic","Definition: Passenger traffic","Passenger traffic is the product of the number of travelers and the distance traveled. The unit is the traveler.km which corresponds to the movement of a traveler over 1km.")+" average of travelers in travelers.km.<br>",o+=a(1)+` Another city bus carries on average $${$t}$ people at a time.<br> The average trip length is $${E}$ km.<br> Show that the passenger traffic is the same as in the question`+a(0)+".",e=a(0)+` The average traffic of this city bus is: $${rt}\\text{ travelers}\\times${K}\\text{ km}=${rt*K}\\text{ travelers.km}$.<br>`,e+=a(1)+` The average traffic of this city bus is: $${$t}\\text{ travelers}\\times${E}\\text{ km}=${$t*E}\\text{ travelers.km}$, so these two buses have the same traffic.`;break;case 6:f=$(0,3),s=$(0,3,[f]),nt=Math.round(H[f][1]/230)+1,o="The devices in this exercise operate on the mains, at a voltage of 230V<br>"+a(0)+` A ${H[f][0]} is protected by a fuse of $${nt}$ amps.<br>What is the`+y(k+p*3+1,w,"power","Definition: Power (physical quantity)","It is the product of the electromotive force (voltage) expressed in Volt (V) by the intensity of the electric current expressed in amperes (A).<br>The unit of measurement of power is the Watt ( W).")+" maximum of this device?<br>",o+=a(1)+` A ${H[s][0]} operates at a maximum power of $${C(H[s][1],0)}$ W.<br>What is the minimum (integer) amperage required for the fuse that will protect ${s===3?"this":"This"} ${H[s][0]} from short circuits?<br>`,e=a(0)+` The mains voltage being $${t(230)}$ V, the maximum power of this ${H[f][0]} is:<br>`,e+=`$230\\text{ V}\\times${nt}\\text{ A}=${t(230*nt)}\\text{ W}$.<br>`,wt=Math.round(H[s][1]/230)+1,e+=a(1)+"To operate at maximum power, this device requires a current with an intensity of:<br>",e+=`$\\dfrac{${C(H[s][1],0)}\\text{ W}}{230 \\text{ V}} = ${t(H[s][1]/230,1)}\\text{ A}$.<br>`,e+=`The fuse required to protect this device from short circuits must have a minimum breaking current of $${wt}$ amps.`;break;case 7:switch(i=Tt[gt],gt++,l=pt(),i){case 0:s=$(0,4),h=$(J[s][1],J[s][2]),n=h*3.6*J[s][3]*$(5,20)/10,r=$(2,J[s][3]),o=`${l} moves ${J[s][0]} to the`+y(k+p*3,w,"speed","Definition: Speed (physical quantity)","Speed is the quotient of the distance traveled by the travel time.<br>The official unit is meters per second ($\\text{m/s}$ or $\\text{m.s}^{- 1}$) but we often use kilometers per hour ($\\text{km/h}$ or $\\text{km.h}^{-1}$).")+` of $${h}\\text{ m/s}$.<br>`,o+=a(0)+` Moving at this speed for $${r}\\text{ h}$, what is the distance covered by ${l}?<br>`,o+=a(1)+` If ${l} wants to travel $${t(n,2)}\\text{ km}$ at this speed, how long will the journey take? Give the result in hours, minutes and seconds.`,e=a(0)+` The distance traveled by ${l} ${J[s][0]} in $${r}$ h at the speed of $${h}\\text{ m/s}$ is:<br>`,e+=`$${t(h,0)}\\text{ m/s}\\times${r}\\text{ h}=\\dfrac{${t(h,0)}\\text{ m}}{1 \\text{ s}}\\times ${r}\\times ${t(3600,0)}\\text{s}`,e+=`=${t(h*3600*r,0)}\\text{ m}=${t(h*3.6*r)}\\text{ km}$.<br>`,e+=a(1)+` To travel $${t(n)}\\text{ km}$ at this speed, let's find the time it will take ${l}.<br>`,e+=` Let's start with the formula $\\mathcal{V}=\\dfrac{\\mathcal{d}}{\\mathcal{t}}$ and replace: $\\dfrac{${h}\\text{ m}}{1 \\text{ s}}=\\dfrac{${t(n)}\\text{ km}}{\\mathcal{t}\\text{ h}}$.<br>`,e+=`Let's make the units homogeneous: $\\dfrac{${h}\\text{ m}}{1 \\text{ s}}=\\dfrac{${C(n*1e3,0)}\\text{ m}}{\\mathcal{t}\\text { h}\\times ${C(3600,0)}\\text{ s/h}}$.<br>`,e+=`Let's apply cross-product equality: ${Ct([[`${h}\\text{ m}`,"1 \\text{ s}"],[`${C(n*1e3,0)}\\text{ m}`,`\\mathcal{t}\\times ${C(3600,0)}\\text{ s/h}`]])}.<br>`,e+=`Hence: $\\mathcal{t}=\\dfrac{1 \\text{ s}\\times${C(n*1e3,0)}\\text{ m}}{${h}\\text{ m}\\times${t(3600)}\\text{ s/ h}}$ ($t$ is the decimal number of hours: meters and seconds disappear because they are both present in the numerator and denominator).<br>`,e+=`Or: $\\mathcal{t}=${t(n*1e3/h/3600,1)}\\times ${C(3600,0)}\\text{ s}=${t(n*1e3/h,0)}\\text{ s}=`,c=Math.floor(n*1e3/h/3600),x=Math.floor(Math.floor(n*1e3/h)%3600/60),g=Math.round(n*1e3/h-3600*c-60*x),c>0&&(e+=`${c}\\times ${C(3600,0)}+`),x>0&&(e+=`${x}\\times 60`),g>0&&(e+=`+${g}`),e=e.replace("++","+")+"\\text{s}=",c!==0&&(e+=`${t(c)}\\text{ h}`),x!==0&&(e+=` ${t(x)}\\text{ min}`),g!==0&&(e+=` ${g}\\text{s}`),e+="$.";break;case 1:r=$(2,15),n=$(5,15,[r])*340,o="Sound travels through the air"+y(k+p*3,w,"speed","Definition: Speed (physical quantity)","Speed is the quotient of the distance traveled by the travel time.<br>The official unit is meters per second ($\\text{m/s}$ or $\\text{m.s}^{- 1}$) but we often use kilometers per hour ($\\text{km/h}$ or $\\text{km.h}^{-1}$).")+" of $340\\text{ m/s}$.<br>",o+=a(0)+` ${l} sees lightning in the sky and counts $${r}$ seconds in his head before he hears thunder.<br>`,o+=`How far from ${l} did the lightning strike?<br>`,o+=a(1)+" The next lightning strike falls on the lightning rod located on the bell tower of the church in the neighboring village.<br>",o+=`${l} knows that the bell tower is located $${t(n)}$ m from its position. How long does it take before ${l} hears thunder?`,e=a(0)+" Let's calculate the distance at which the first lightning bolt fell using the speed of sound (we consider that the speed of light is such that the lightning bolt is visible instantly):<br>",e+=`$340\\text{ m/s}=\\dfrac{340\\text{ m}}{1\\text{ s}}=\\dfrac{${ot(r)}\\times 340\\text{ m}}{${ot(r)} \\times 1\\text{ s}}=\\dfrac{${t(r*340,0)}\\text{ m}}{${r}\\text{ s}}$<br>`,e+=`The distance at which the lightning fell is therefore $${t(r*340,0)}\\text{ m}$.<br>`,e+=a(1)+" With the data from the statement, we can write:<br>",e+=`$\\dfrac{340\\text{ m}}{1\\text{ s}}=\\dfrac{${t(n,0)}\\text{ m}}{\\mathcal{T}\\text{ s}}$ .<br>`,e+=`Or thanks to the equality of the cross products: $\\mathcal{T}\\text{ s}=${It("340 \\text{ m}","1 \\text{ s}",t(n,0)+"\\text{ m}",0)}=${t(n/340,0)}\\text{ s}$.<br>`,e+=`${l} will hear the thunder $${t(n/340,0)}$ seconds after seeing the lightning strike the bell tower.`;break;case 2:h=$(J[4][1]*5,J[4][2]*5)/5,n=$(5,12),l=lt(),o=`${l} just ran ${n} kilometers. His smartwatch recorded the`+y(k+p,w,"allure","Definition: Pace (physical quantity)","Pace is the time expressed in h, min, s to cover one kilometer.<br>The unit is then h/km or min/km.")+"for each kilometer traveled:",m=[];for(let q=0;q<n;q++)r=Math.round(1e3/(h*(1+$(-10,10)*.01))),g=r%60,x=(r-g)/60,m.push([x,g]);o+="$\\def\\arraystretch{1.5}\\begin{array}{|c",o+="|c";for(let q=0;q<m.length;q++)o+="|c";o+="}\\hline \\text{kilometer}";for(let q=0;q<m.length;q++)o+="&"+t(q+1);for(o+="\\\\\\hline \\text{pace in minutes and seconds (per km)}",st=0;st<m.length;st++)o+="&"+m[st][0]+"\\text{ min }"+m[st][1]+"\\text{s}";o+="\\\\\\hline\\end{array}$<br>",o+=a(0)+` Calculate the total duration of the ${l} race.<br>`,o+=a(1)+" Deduce its"+y(k+p+1,w,"speed","Definition: Speed (physical quantity)","Speed is the quotient of the distance traveled by the travel time.<br>The official unit is meters per second ($\\text{m/s}$ or $\\text{m.s}^{- 1}$) but we often use kilometers per hour ($\\text{km/h}$ or $\\text{km.h}^{-1}$).")+" average in $\\text{km/h}$ over the total journey.<br>",o+=a(2)+` ${l} is training for a half marathon ($${t(21.0975,4)}\\text{ km}$). Running at the same average speed, how long would your half marathon last?`,e=a(0)+` The total duration of the ${l} race is:<br>`,m.push([0,0]),r=0;for(let q=0;q<n;q++)m[n][1]+=m[q][1],m[n][1]>59&&(m[n][0]+=1,m[n][1]=m[n][1]%60),m[n][0]+=m[q][0],m[n][0]>59&&(r++,m[n][0]=m[n][0]%60);for(let q=0;q<n-1;q++)e+=`$${m[q][0]}\\text{ min }${m[q][1]}\\text{ s }+${V()}$`;e+=`$${m[n-1][0]}\\text{ min }${m[n-1][1]}\\text{ s }= ${V()}$`,r!==0&&(e+=`$${r}\\text{ h}$`),m[n][0]!==0&&(e+=`$${V()}${m[n][0]}\\text{ min}$`),m[n][1]!==0&&(e+=`$${V()}${m[n][1]}\\text{ s}$.`),e+="<br>"+a(1)+` ${l} performed $${n}\\text{ km}$ in`,r!==0&&(e+=`$${r}\\text{ h}$`),m[n][0]!==0&&(e+=`$${V()}${m[n][0]}\\text{ min}$`),m[n][1]!==0&&(e+=`$${V()}${m[n][1]}\\text{ s}$.<br>Let`),r!==0&&(e+=`$${r}\\text{ h}$`),m[n][0]!==0&&(e+=` $\\dfrac{${m[n][0]}}{60}\\text{ h}$`),m[n][1]!==0&&(e+=` $\\dfrac{${m[n][1]}}{${t(3600)}}\\text{ h}$ =`),e+="$\\dfrac{",r!==0&&(e+=`${r}\\times ${t(3600)} +`),e+=`${m[n][0]}\\times 60+${m[n][1]}}{${t(3600)}}\\text{ h}$ =`,e+="$\\dfrac{",r!==0?(r=r*3600+m[n][0]*60+m[n][1],e+=`${t(r)}}`):(r=m[n][0]*60+m[n][1],e+=`${t(r)}}`),e+=`{${t(3600)}}\\text{ h}$.<br>`,e+=`Its average speed in $\\text{km/h}$ is therefore:<br>$${n} \\text{ km}\\div\\dfrac{${r}}{${t(3600)}}\\text{ h}=`,e+=`${n} \\text{ km}\\times\\dfrac{${t(3600)}}{${r}}\\text{ h}^{-1}=\\dfrac{${n}\\times${t(3600)}}{${r}}\\text{km.h }^{-1}`,h=Number((n*3600/r).toFixed(1)),e+=`\\approx${t(h,1)}\\text{ km/h}$.<br>`,e+=a(2)+` If she runs $${t(21.0975,4)}\\text{ km}$ at this average speed of $${t(h,1)}\\text{ km/h}$, ${l} will put:<br>`,r=Number((21.0975/h).toFixed(4)),e+=`$\\dfrac{${t(21.0975)} \\text{ km}}{${t(h,1)} \\text{ km.h}^{-1}}\\approx${t(r,4)}\\text{ h}$, i.e.`,c=Math.floor(r),e+=`$${t(r,4)}\\times ${C(3600,0)}\\text{ s}\\approx${t(r*3600,0)}\\text{ s}=`,r=(Number(r.toFixed(4))-c)*60,x=Math.floor(r),r=Math.round((Number(r.toFixed(3))-x)*60),g=r,c>0&&(e+=`${c}\\times ${C(3600,0)}+`),x>0&&(e+=`${x}\\times 60`),g>0&&(e+=`+${g}`),e=e.replace("++","+")+"\\text{s}=",c!==0&&(e+=`${t(c)}\\text{ h}`),x!==0&&(e+=` ${t(x)}\\text{ min}`),g!==0&&(e+=`${g}\\text{s}`),e+="$.";break}break;case 8:s=$(0,7),i=$(0,5,[s]),f=$(0,5,[s,i]),d=$(u[s][2],u[s][3])/10,O=$(u[i][2],u[i][3])/10,mt=$(u[f][2],u[f][3])/10,Y=d*u[s][1],B=O*u[i][1],ut=mt*u[f][1],l=lt(),o=`${l} goes to his neighborhood grocery store. She buys $${t(d)}\\text{ kg}$ of ${u[s][0]} at $${v(u[s][1])}$ €$\\text{/kg }$ and for $${v(B)}$ € of ${u[i][0]} at $${v(u[i][1])}$ €$\\text{/kg }$ .<br>`,o+=`Finally, she buys $${t(mt)}\\text{ kg}$ of ${u[f][0]} for $${v(ut)}$ €.<br>`,o+=a(0)+` How much does ${u[s][0]} cost him?<br>`,o+=a(1)+` How much ${u[i][0]} did she buy${u[i][0]==="lemons"?"s":"are"}?<br>`,o+=a(2)+` What is the price per kilogram of ${u[f][0]}?`,e=a(0)+` ${l} expenditure for ${u[s][0]}: $${t(d)}\\text{ kg}\\times ${v(u[s][1])}$ €$\\text{/kg }=${v(Y)}$ €.<br>`,e+=a(1)+` The mass of ${u[i][0]} she purchased${u[i][0]==="lemons"?"s":"are"} is: $${v(B)}$ € $~\\div~ ${v(u[i][1])}$ €$\\text{/kg } = ${t(O)}\\text{ kg}$.<br>`,e+=a(2)+` Finally, ${l} purchased ${u[f][0]}s at the unit price of: $${v(ut)}$ € $~\\div~ ${t(mt)}\\text{ kg } = ${v(u[f][1])}$ €$\\text{/kg}$.`;break;case 9:s=$(0,3),i=$(0,4),c=$(it[s][2],it[s][3]),Y=it[s][1],B=Z[i][1]*$(2,6),l=lt(),o=`${l} has planned to rent ${it[s][0]} for $${t(c)}$ hours. The rental hour costs $${v(Y)}$ €.<br>`,o+=a(0)+" How much will this rental cost him?<br>",o+=a(1)+` ${l} took private ${Z[i][0]} lessons. In total, this month, she had $${C(B/Z[i][1],0)}$ hours of lessons for $${v(B)}$ €. How much does his teacher charge for an hour of lessons?<br>`,e=a(0)+` ${l} will spend for its ${it[s][0]} rental: $${t(c)}\\text{ h} \\times ${v(Y)}$ €$\\text{/h} = ${v(c*Y)}$ €.<br>`,e+=a(1)+` The ${Z[i][0]} lesson hour costs: $${v(B)}$ € $${V()}\\div${V()}${t(B/Z[i][1])}\\text{ h} = ${v(Z[i][1])}$ €$\\text{/h}$.<br>`;break;case 10:s=$(0,14),i=$(0,14,[s]),o=a(0)+` In 2016, in ${b[s][0]} there were $${t(b[s][1])}$ inhabitants for an area of $${t(b[s][2]*100,0)}\\text{ ha}$.<br> Calculate the population density in $\\text{hab/km}^2$ rounded to the nearest unit.<br>`,o+=a(1)+" The same year, the"+y(k+p*3+1,w,"population density","Definition: Population density","It is the quotient of the number of inhabitants by the surface area in km$^2$.<br>The unit of population density is the inhabitant per km$^2$ (inhabitant/km$ ^2$).")+` of ${b[i][0]} was $${t(b[i][1]/b[i][2],0)}\\text{ hab/km}^2$ for an area of $${t(b[i][2]*100,0)}\\text{ ha}$.<br> Calculate the number of inhabitants of ${b[i][0]} on this date.<br>`,e=a(0)+` In 2016, the population density in ${b[s][0]} was:<br> $\\dfrac{${t(b[s][1],0)}\\text{ hab}}{${t(b[s][2]*100,0)}\\text{ ha}}=\\dfrac{${t(b[s][1],0)}\\text{ hab}} {${t(b[s][2],2)}\\text{ km}^2}\\approx${t(b[s][1]/b[s][2],0)}\\text{ hab/km}^{2}$.<br>`,e+=a(1)+` On this date, the number of inhabitants of ${b[i][0]} was:<br> $${t(b[i][1]/b[i][2],0)}\\text{ hab/km}^2\\times ${t(b[i][2]*100,0)}\\text{ ha}=${t(b[i][1]/b[i][2],0)}\\text{ hab/km}^ 2\\times ${t(b[i][2],1)}\\text{ km}^{2}\\approx${t(b[i][1],0)}\\text{ hab}$.`;break;case 11:s=$(0,14),i=$(0,14,[s]),tt=$(50,100),O=$(5,30),d=W[s][1]*tt/1e6,et=O/W[i][1],o=a(0)+" The "+y(k+p*3+1,w,"volumic mass","Definition: Density (physical quantity)","The density of an element is the quotient of the mass of this element by the volume it occupies.<br>The unit of density depends on the nature of the element and can be expressed in kg/m$^3$ for solids or in g/L for gases, for example.")+` ${W[s][2]}${W[s][0]} is $${t(W[s][1])}\\text{ kg/m}^3$.<br>`,o+=`What is the mass of a piece of this metal of $${t(tt)}\\text{ cm}^3$?<br>`,o+=a(1)+` What is the volume of a ${W[i][3]}${W[i][0]} part having a mass of `,o+=`$${t(O)}\\text{ kg}$ (the ${W[i][2]}${W[i][0]} density is $${t(W[i][1])}\\text{ kg/m}^3$)?<br>`,e=a(0)+` The mass of this ${W[i][3]}${W[s][0]} part is:<br>$${t(W[s][1],0)}\\text{ kg/m}^3\\times ${t(tt,0)}\\text{ cm}^3=${t(W[s][1],0)}\\text{ kg/m}^3\\ times ${t(tt/1e6,6)}\\text{ m}^3=${t(d,6)}\\text{ kg}${bt(d*1e3,0)}${t(d*1e3,0)}$ g.<br>`,e+=a(1)+` The volume of this ${W[i][3]}${W[i][0]} part is:<br>$${t(O,0)}\\text{ kg}\\div ${t(W[i][1],0)}\\text{ kg/m}^3${bt(et,6)}${t(et,6)}\\text{ m}^3${bt(Dt(et,6)*1e6,0)}${t(et*1e6,0)}\\text{ cm}^3 $.<br>`;break;case 12:s=$(0,4),i=$(0,4,[s]),R=$(2,15,[10])/10,ht=$(2,15,[10])/10,F[s][2]<10?d=$(11,F[s][2]*10)*R/10:d=$(2,F[s][2])*R,F[i][2]<10?_=$(11,F[i][2]*10)/10:_=$(2,F[i][2]),o="The mass concentration expressed here in $g/L$ is the quantity of material (mass) in g per unit of volume (L).<br>It is therefore obtained by dividing the mass of dissolved product by the volume of the solution ($\\dfrac{m}{V}$).<br>"+a(0)+` We dissolved $${t(d,1)}\\text{ g}$ of ${F[s][0]} in $${t(R,1)}\\text{ liters}$ ${F[s][1]}.<br>Calculate the mass concentration of this solution.<br>`,o+=a(1)+` We have $${t(ht,1)}$ liters of aqueous solution of ${F[i][0]} at $${t(_,1)}\\text{ g/L}$.<br>What mass of ${F[i][0]} was dissolved in water?`,e=a(0)+` The ${F[s][0]} concentration of this aqueous solution is:<br>`,e+=` $\\dfrac{${t(d,1)}\\text{ g}}{${t(R,1)}\\text{ L}}=${t(d/R,1)}\\text{ g/L}.$<br>`,e+=a(1)+` The mass of dissolved ${F[i][0]} is:<br>`,e+=`$${t(ht,1)}\\text{ L}\\times ${t(_,1)}\\text{ g/L}=${t(ht*_,2)}\\text{ g}.$`;break;case 13:i=$(0,6),r=$(2,24),U=T[i][3]*3600,o="THE"+y(k+p,w,"Speed","Definition: Flow (physical quantity)","The flow rate is the quotient of a volume of water flowing in a section of conduit by the flow time.<br>The official unit is the cubic meter per second ($\\text{m} ^3/\\text{s}$) and in some cases, liter per minute (L/min) can be used.")+` Average annual ${T[i][6]}${T[i][0]} measured at ${T[i][1]} is $${t(T[i][2])}\\text{ m}^3\\text{/s}$.<br>`,o+=a(0)+` Calculate the volume of water in m${at(3)} flowed in $${r}$ hours at this flow rate.<br>`,o+=a(1)+` In ${T[i][4]} to ${T[i][1]}, ${T[i][5]}${T[i][0]} debited $${t(U)}\\text{ m}^3$ in one hour. What was then its flow rate in $\\text{ m}^3\\text{/s}$?`,e=a(0)+` In $${r}$ hours, on average it passes in ${T[i][5]}${T[i][0]} to ${T[i][1]}:<br>`,e+=`$\\mathcal{V}=${r}\\text{ h}\\times${T[i][2]}\\text{ m}^3\\text{/s}=${r}\\times 3600\\text{ s}\\times${T[i][2]}\\ text{ m}^3\\text{/s}=${t(r*3600*T[i][2])}\\text{ m}^3$.<br>`,e+=a(1)+` In ${T[i][4]}, during the historic flood ${T[i][6]}${T[i][0]} to ${T[i][1]}, the flow was:<br>`,e+=`$\\text{Flow} = ${t(U)}\\text{ m}^3\\text{/h}=\\dfrac{${t(U)}\\text{ m}^3}{1\\text{ h}}=\\dfrac{${t(U)}\\text{ m}^3}{${t(3600)}\\text{ s}}=${t(U/3600,0)}\\text{ m}^3\\text{/s}$.<br>`;break;case 14:j=["ko","MB","Go"],f=$(0,1),f===0?h=$(200,999):h=$(1,20),l=pt(),x=$(3,10),g=$(2,59),d=$(15,35)/10,o=a(0)+` ${l} downloads a file from online storage. Her `+y(k+p,w,"download speed","Definition: Download Speed","The download speed is the quotient of the quantity of data downloaded (in KB, MB or GB) by the download duration (in seconds).<br>The unit of this quotient quantity is the KB/s (or MB /s)")+` is $${h}$ $\\text{${j[f]}/s}$.<br>`,o+=`The download takes $${x}$ minutes and $${g}$ seconds. What is the size of the downloaded file in $\\text{${j[f]}}$?<br>`,o+=a(1)+` ${l} wants to download a file of $${t(d,1)}\\text{ GB}$. How long will the download take if its download speed is $${h}$ $\\text{${j[f]}/s}$?<br>`,e=a(0)+" The size of the downloaded file is:<br>",P=(x*60+g)*h,e+=`$(${x}\\times 60 +${g})\\text{ s}\\times ${h} \\text{ ${j[f]}/s} = ${x*60+g}\\text{ s}\\times ${h} \\text{ ${j[f]}/s} = ${t(P,0)} \\text{ ${j[f]}}$`,P>1e3&&(e+=`$ =${t(P/1e3,3)} \\text{ ${j[f+1]}}$`),e+=".<br>",e+=a(1)+" The download duration will be:<br>",f===0?(e+=`$${t(d,1)}\\times ${t(10**6,0)} \\text{ kb} \\div ${h} \\text{ ${j[f]}/s}$`,P=d*10**6):(e+=`$${t(d,1)}\\times ${t(10**3,0)} \\text{ MB} \\div ${h} \\text{ ${j[f]}/s}$`,P=d*10**3),e+=`$${V(2)}=\\dfrac{${t(P,0)}}{${h}}\\text{ s}`,c=Math.floor(P/h/3600),x=Math.floor((P/h-3600*c)/60),g=Math.floor(P/h-3600*c-60*x),P/h===g+60*x+3600*c?e+="=":e+="\\approx",c!==0&&(e+=`${c} \\text{ h }`),x!==0&&(e+=`${x} \\text{ min }`),g!==0&&(e+=`${g} \\text {s}`),e+="$.";break}this.listeQuestions.indexOf(o)===-1&&(this.listeQuestions.push(o),this.listeCorrections.push(e),p++),yt++}Ft(this)},this.besoinFormulaireTexte=["Type of quantities",`Numbers separated by hyphens
 1: Energy consumed
 2: Volume
 3: Quantity of movement & Kinetic energy
 4: Moment of force
 5: Passenger traffic
 6: Electric power
 7: Speed
 8: Mass price
 9: Hourly price
 10: Population density
 11: Density
 12: Mass concentration
 13: Flow rates
 14: File transfer`]}export{zt as default,Bt as ref,St as titre,Qt as uuid};
//# sourceMappingURL=4P10-Z69Btd8I.js.map