Дано: V = 2 × 10⁻² м³, ΔT = 91 K, T = 273 K, p = 10⁵ Па, i = 3
Найти: Q - ?
Решение:
![\boxed{pV = \nu RT} \ ; \ \nu = \dfrac{pV}{RT} = \dfrac{10^{5}*2*10^{-2}}{8.31*273} \approx 0.88 \ [mol] \\ \\ \boxed{\Delta U = Q + A_H_._\Gamma} \ ; \ \rightarrow \ \Delta U = Q \ (\Delta V = 0) \ ; \ \Delta U = \dfrac{i}{2}\nu R\Delta T \\ \\ Q = \dfrac{3}{2}\nu R\Delta T = 1.5 * 0.88*8.31*91 = 998 \ [Ds] \approx 1000 \ [Ds]](https://tex.z-dn.net/?f=%5Cboxed%7BpV+%3D+%5Cnu+RT%7D+%5C+%3B+%5C+%5Cnu+%3D++%5Cdfrac%7BpV%7D%7BRT%7D+%3D++%5Cdfrac%7B10%5E%7B5%7D%2A2%2A10%5E%7B-2%7D%7D%7B8.31%2A273%7D+%5Capprox+0.88+%5C+%5Bmol%5D+%5C%5C+%5C%5C+%5Cboxed%7B%5CDelta+U+%3D+Q+%2B+A_H_._%5CGamma%7D+%5C+%3B+%5C+%5Crightarrow+%5C+%5CDelta+U+%3D+Q+%5C+%28%5CDelta+V+%3D+0%29+%5C+%3B+%5C+%5CDelta+U+%3D++%5Cdfrac%7Bi%7D%7B2%7D%5Cnu+R%5CDelta+T+%5C%5C+%5C%5C+Q+%3D+%5Cdfrac%7B3%7D%7B2%7D%5Cnu+R%5CDelta+T+%3D+1.5+%2A+0.88%2A8.31%2A91+%3D+998+%5C+%5BDs%5D+%5Capprox+1000+%5C+%5BDs%5D)
Ответ: Q = 1000 Дж
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16м/с=60*16=960м/мин=960*60=57600 м/ч=57.6км/ч
160км/ч>57.6км/ч
Следовательно ответ,нет,не сможет.
f = 3d
3d = 4F
F =
= 40 см = 0,4 м
f = 3d = 3 * 40 = 120 см = 1,2 м
1,2 + 0,4 = 1,6 м
вариант 5
U1=U2=60 B
I1=U1/R1=60/6=10 A
I2=U2/R2=60/12=5A
I3=I4=(I1+I2)R5/(R3+R4+R5)=(10+5)12/(10+2+12)=15·12/24=7,5
U3=I3·R3=7,5·10=75 B
U4=I4·R4=7,5·2=15 B
U5=U3+U4=75+15=90 B
I5=U5/R5=90/7,5=12 B
Вариант 6
R(1-3)=R1·R2/(R1+R2)+R3=3·6/(3+6)+2=2+2=4 Ом
R(4-5)=R4·R5/(R4+R5)=12·6/(12+6)=72/18=4 Ом
R=R(1-3)·R(4-5)/[R(1-3)+R(4-5)]=4·4/(4+4)=2 Ом
I=Uаб/R=18/2=9 В
