При индуктивности катушки 0, 1 мгн резонансная частота цепи lc была равна 10 кгц. какова будет резонансная частота цепи при индуктивности 0, 2 мгн ? (решите это

L1=0.1*10^-3 Гн    v1=10*10^3 Гц     L2=0.2*10^-3 Гн    v2=?
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v1/v2=√(L2/L1)
v2=v1*√(L1/L2)=10*10^3*√(0.1/0.2)=7.1*10^3 Гц

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n=19%

Qотд=53 МДж

Объяснение:

Для начала переведём все в систему СИ:

t=4 часа= 14400 с

V=2 л=0.002 м^3

Обозначим КПД за n. Тогда:

n=P*t/(q*m)

где P - мощность ДВГ, t - время его работы, q=4.6*10^7 - удельная теплота сгорания бензина, m - масса бензина.

Массу бензина можно найти по формуле:

m=ρ*V

где ρ=710 кг/м^3 - плотность бензина, V - его объём.

подставив всё в формулу получим:

n=P*t/(q*ρ*V)=860*14400/(4.6*10^7*710*0.002)=0.1895=19%

Найдём теплоту отданную холодильнику из формулы:

n=A/Q1; A=Q1-Qотд, где Qотд - отданная холодильнику теплота (по модулю).

A=n*Q1; n*Q1=Q1-Qотд   =>   Qотд=Q1-n*Q1=Q1*(1-n)=q*ρ*V*(1-n)=4.6*10^7*710*0.002*(1-0.1895)=52941860=53 МДж

Всё... Фух.

:)