помогите решить задачу пож-та)))
сила тока в электросварочном аппарате в момент сварки 7500А при напряжении 3 В.свариваемые стальные листы при этом имеют сопротивление 0,0004 Ом.Какое количество теплоты выделяется при сварке за 2 минуты?
Дано:
I=7500 А,
U=3 В.
R=0,0004 Ом.
t=2 м=120 с.
Q=?______Решение:Применяем закон Джоуля Ленца:
Можно брать любую из трех формулировок, все подойдут. Мы возьмем какая проще -1ую.
Q=7500*3*120=2700000 Дж=2,7 МДж.
Ответ: Q=2,7 МДж.
Автор:
cubbywongПо формуле:
Q= J^2*R*t
Q= 7500^2 A* 0.0004 Ом* 120 c= 2700000 Дж .
Ответ: 2,7 МДж .
Автор:
garrett718Добавить свой ответ
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