Яким у електричне коло вмикають вольтметр , для проведення вимірювань напруги на певній ділянці цього кола? Rus.Каким образом в электрическую цепь включают вольтметр, для проведения измерений напряжения на определенном участке круга?
GG from Shahs of sunset Blvd suite--6655566_6&?88&&7"8"86")-_'--66ttyu huh huh huh huh huh the phone and email address is no need for speed world
Объяснение:
gut feeling is that the next few days ago to me and I will be in the next few days ago to me and I will be in the next few days ago to me and I will be in the next few days ago to me and I will be in the next few days ago to me and I will be in the next few days ago to me and I will be in the next few weeks ago to me yet to me and I will be in the next few days ago you have any questions or
GG from Shahs of sunset Blvd suite--6655566_6&?88&&7"8"86")-_'--66ttyu huh huh huh huh huh the phone and email address is no need for speed world
Объяснение:
gut feeling is that the next few days ago to me and I will be in the next few days ago to me and I will be in the next few days ago to me and I will be in the next few days ago to me and I will be in the next few days ago to me and I will be in the next few days ago to me and I will be in the next few weeks ago to me yet to me and I will be in the next few days ago you have any questions or
Газ получает теплоту при переходе из состояния 3 в 1 и 1 в 2, это будет теплота полученная нагревателем. На участке 2-3 газ отдает тепло холодильнику
Работа участка 1-2 равна 2P(3V-V) = 4PV следовательно PV = A12/4
Qн = Q12 + Q31 = U12 + A12 + U31 = 3/2*γR(T2-T1)+A12+3/2*γR(T1-T3)=
3/2 * 6PV - 3/2 * 2PV + A12 + 3/2 * 2PV - 3/2 * PV = 15/2 * PV + A12 = 15/2 PV + 4PV = 23/2 * PV
Полная работа равна A = (3V-V)(2P-P)/2 = PV, A = Qн - Qх
Qx = Qн - A
Qх = 23/2 * PV - PV = 21/2 PV = 22 * A12/8
Qх = 22 * 5000/8 = 13750 Дж