R13=R1+R3=8+2=10 Ом. R24=R2+R4=10+5=15 Ом. Искомое Rоб=R13*R24/(R13+R24)=10*15/(10+15)=150/25=6 Ом.
Ответ: 6 Ом.
1)A=Q
A=nhv v=c/лямда
Q=cm(t2-t1)=42*10^3 Дж
A=10^20*6.63*10^-34*3*10^8/330*10^-9=6*10^-7
t=Q/A=7*10^10 c
V=0.6 м^3, Fa=5 кН=5000 Н, ρ(v)=1000 кг/м^3;
V=V(nad)+V(pod)
Fa=ρ*g*V =>
![V_{pod}=\frac{F_{a}}{\rho_{v}*g}=\frac{5000}{1000*10}=0.5](https://tex.z-dn.net/?f=V_%7Bpod%7D%3D%5Cfrac%7BF_%7Ba%7D%7D%7B%5Crho_%7Bv%7D%2Ag%7D%3D%5Cfrac%7B5000%7D%7B1000%2A10%7D%3D0.5)
V(nad)=V-V(pod)=0.6-0.5=0.1 м^3