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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
1. mv^2/2=mgh; h=144/20=7.2 м
2. m1v1=m2v2; v1=18*1700/19=1610 м/c
3. Eк=Еп, т.к. в в момент касания Еп=0; Eк=mgh=0,04*2*10=0.8 Дж
4. F=GMm/(R+H)^2; F= 4,54*10^12
Объяснение: