айдем сопротивление нитей ламп, зная мощность и напряжение принимаем 220 вольт, P=U*I если I=U/R то подставив получим P=U* U/R=U²/R отсюда R1=U²/P=220²/40=48400/40=1210 Ом -сопротивление 1 лампы
R2=U²/P=220²/60=48400/60=807 Ом - сопротивление 2 лампы
лампы будут включены последовательно, значит общее сопротивление цепи равно R=R1+R2=1210+807= 2017 Ом сопротивление всей цепи, в последовательной цепи ток одинаковый значит I=U/R=220/2017=0.11 А, зная ток и сопротивление находим мощность каждой лампы P=I*U если I=U/R то U=I*R подставим в выражение P=I*I*R1=I²*R1= 0.0121*1210 =14.64 Вт - первая лампа
P=I*I*R2=I²*R2= 0.0121*807=9.76 вторая лампа
проверяем
U1=I*R1= 0.11*1210=133.1 Вольт - напряжение на первой лампе U2=I*R2=0.11*807=88.7 - напряжение на второй лампе ( в последовательной цепи общее напряжение равно сумме напряжений на каждом сопротивлении ) P1=I*U1=0.11*133.1=14.64 Вт - 1 лампа, P2=I*U2=0.11*88.7=9.76 Вт -2 лампа
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ответ: 40284 Дж или 40,3 кДж
Объяснение:
Формула плавления Q=λm
Формула нагрева Q=cm(t2-t1)
Q общее = Q1+Q2
Дано:
m= 200г = 0,2кг(СИ)
λ= 0,67x
tпл золота = 1064 гр.C
c золота = 130 Дж/кгxгр.C
Q1 = 130x0,2(1064-30) = 26884 Дж. (Столько энергии потратится чтобы золото нагрелось до температуры плавления)
Q2 = (0,67x
) x 0,2 = 0,134x (Столько энергии будет потрачено при плавлении уже нагретого золота)
Q общее = 26884+13400 = 40284 Дж. = 40.3 кДж
(X значит умножение)