1.
R = 50 см = 0,5 м
t = 3,14 c
![v-?](https://tex.z-dn.net/?f=v-%3F)
![v= \frac{2 \pi R}{T}](https://tex.z-dn.net/?f=v%3D+%5Cfrac%7B2+%5Cpi+R%7D%7BT%7D+)
,
![T=2t](https://tex.z-dn.net/?f=T%3D2t)
![v= \frac{ \pi R}{t}](https://tex.z-dn.net/?f=v%3D+%5Cfrac%7B+%5Cpi+R%7D%7Bt%7D+)
![v= \frac{3,14*0,5}{3,14} =0,5](https://tex.z-dn.net/?f=v%3D+%5Cfrac%7B3%2C14%2A0%2C5%7D%7B3%2C14%7D+%3D0%2C5)
м/с
2.
R = 2 м
![v=0,5](https://tex.z-dn.net/?f=v%3D0%2C5)
м/с
ω - ?
![v= \frac{ \pi R}{t}](https://tex.z-dn.net/?f=v%3D+%5Cfrac%7B+%5Cpi+R%7D%7Bt%7D+)
,
![T= \frac{2 \pi R}{v}](https://tex.z-dn.net/?f=T%3D+%5Cfrac%7B2+%5Cpi+R%7D%7Bv%7D+)
ω =
ω =
ω =
![\frac{0,5}{2}=0,25](https://tex.z-dn.net/?f=+%5Cfrac%7B0%2C5%7D%7B2%7D%3D0%2C25+)
(рад/с)
3.
ω = 0,25 рад/с
t = 1 мин = 60 с
ν - ?
ω = 2πν
ν = ω/2π
ν = 0,25/2*3,14=3,98 об/с = 239 об/мин
4.
ν = 239 об/мин
R = 2 м
![v-?](https://tex.z-dn.net/?f=v-%3F)
![v= \frac{2 \pi R}{T}](https://tex.z-dn.net/?f=v%3D+%5Cfrac%7B2+%5Cpi+R%7D%7BT%7D+)
, T = 1/ν
![v=2 \pi R](https://tex.z-dn.net/?f=v%3D2+%5Cpi+R)
ν
![v=2*3,14*2*239=3002](https://tex.z-dn.net/?f=v%3D2%2A3%2C14%2A2%2A239%3D3002+)
м/мин
КПД=Р*100% / Q Q=Р*100%/КПД=15*10^5*100%\30%=5*10^6 Дж
Дано: F = 3.6 H, P = 2.8 H, ρ(ц) = 7100 кг/м³
Найти: Vпол - ?
Решение:
![\vec R = 0 \ ; \ \vec F_A + \vec F_T + \vec P = 0 \\ \\ OY: F_T - P - F_A = 0 \\ \\ F_A = F_T - P = 3.6\ H - 2.8\ H = 0.8 \ [H] \\ \\ F_A = \rho g V_T \ ; \ V_T = \dfrac{F_A}{\rho g} \ ; \ m = \rho V \ ; \ F_T = mg \\ \\ 1)\ V_T = \dfrac{0.8}{1000*10} = 8 * 10^{-5} \ [M^{3}] \\ \\ 2)\ F_T = \rho_{1}Vg \ ; \ V = \dfrac{F_T}{\rho_{1}g} = \dfrac{3.6}{7100*10} = 5.07 * 10^{-5} \ [M^{3}] \\ \\ V_\Pi = V_T - V = 8 * 10^{-5} - 5.07 * 10^{-5} = 2.93 * 10^{-5} \ [M^{3}] \\ \\](https://tex.z-dn.net/?f=%5Cvec+R+%3D+0+%5C+%3B+%5C+%5Cvec+F_A+%2B+%5Cvec+F_T+%2B+%5Cvec+P+%3D+0+%5C%5C+%5C%5C+OY%3A+F_T+-+P+-+F_A+%3D+0+%5C%5C+%5C%5C+F_A+%3D+F_T+-+P+%3D+3.6%5C+H+-+2.8%5C+H+%3D+0.8+%5C+%5BH%5D+%5C%5C+%5C%5C+F_A+%3D+%5Crho+g+V_T+%5C+%3B+%5C+V_T+%3D++%5Cdfrac%7BF_A%7D%7B%5Crho+g%7D+%5C+%3B+%5C+m+%3D+%5Crho+V+%5C+%3B+%5C+F_T+%3D+mg+%5C%5C+%5C%5C+1%29%5C+V_T+%3D++%5Cdfrac%7B0.8%7D%7B1000%2A10%7D+%3D+8+%2A+10%5E%7B-5%7D+%5C+%5BM%5E%7B3%7D%5D+%5C%5C+%5C%5C+2%29%5C+F_T+%3D+%5Crho_%7B1%7DVg+%5C+%3B+%5C+V+%3D++%5Cdfrac%7BF_T%7D%7B%5Crho_%7B1%7Dg%7D+%3D++%5Cdfrac%7B3.6%7D%7B7100%2A10%7D+%3D+5.07+%2A+10%5E%7B-5%7D+%5C+%5BM%5E%7B3%7D%5D+%5C%5C+%5C%5C+V_%5CPi+%3D+V_T+-+V+%3D+8+%2A+10%5E%7B-5%7D+-+5.07+%2A+10%5E%7B-5%7D+%3D+2.93+%2A+10%5E%7B-5%7D+%5C+%5BM%5E%7B3%7D%5D+%5C%5C+%5C%5C+)
Ответ: Vпол = 2.93 × 10⁻⁵ м³
Маляр прикладывает усилие F
сила F действует на руки маляра со стороны веревки
противоположный конец веревки с силой F воздействуют на кресло с человеком
(m+M)*a = 2*F - mg - Mg
a = 2F/(m+M)-g = 2*500/(30+70)-10 м/с^2 = 0 м/с^2 - человек поднимается без ускорения
Работа за цикл
A = Q1 - Q2 = 3 - 2.4 = 0.6 кДж
Затраты за цикл Q1 = 3 кДж
Отношение работы к затратам
η = A/Q1 = 0.6/3 = 0.2
Значит, кпд 20\%