Предмет: Физика, автор: ruslanhd4k

До джерела струму з напругою 6 В на 1,5 хв приєднали два паралельно з’єднаних провідники опорами 30 Ом і 60 Ом. Яка загальна робота була виконана електричним струмом?

Ответы

Автор ответа: geroinggg1
0

R=(R1*R2)/(R1+R2)=1800/90=20 Oм

I=U/R=6/20=0,3 A

Iзагальне=q/t

q=I*t=0,3*90=27 Дж/кг

U=A/q

A=U*q=20*27=540 Дж

Відповідь: 540 Дж

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