Дано: Решение:
R1=1ОМ R1,2=R1+R2=1ом+2ом=3ом
R2=2ОМ R3,4=R3+R4=3ом+4ом=7ом
R3=3ОМ R0=(R1,2*R3,4)/(R1,2+R3,4)=(3ОМ*7ОМ).(3ОМ+7ОМ)=
R4=4ОМ =21ОМ/10ОМ=2,1ом
Найти: Ответ:2,1ом.
R0
<em>P=F/S</em>
<em>F=mg</em>
<em>S=0,4*0,05=0,02см^2</em>
<em>P=(mg/0,02см^2)*2=(80*10/0,02)*2=80000Па=80кПа</em>
<em>Ответ:80кПа</em>
1 способ:
![mgh = \frac{mv^2}{2}](https://tex.z-dn.net/?f=mgh+%3D++%5Cfrac%7Bmv%5E2%7D%7B2%7D+)
------------------->
![v = \sqrt{ \frac{2mgh}{m}](https://tex.z-dn.net/?f=v+%3D++%5Csqrt%7B+%5Cfrac%7B2mgh%7D%7Bm%7D+)
=
![\sqrt{2gh](https://tex.z-dn.net/?f=+%5Csqrt%7B2gh)
g - ускорение свободного падения ( 10 Н/кг )
h - высота ( 25 м )
![v = \sqrt{2*10*25](https://tex.z-dn.net/?f=v+%3D++%5Csqrt%7B2%2A10%2A25)
=
![\sqrt{500](https://tex.z-dn.net/?f=+%5Csqrt%7B500)
≈ 22,36 м/c
2 способ:
h =
![\frac{V^2}{2g}](https://tex.z-dn.net/?f=+%5Cfrac%7BV%5E2%7D%7B2g%7D+)
---------------------->
![V = \sqrt{h * 2g} = \sqrt{2gh} = \sqrt{2*10*25} = \sqrt{500](https://tex.z-dn.net/?f=V+%3D++%5Csqrt%7Bh+%2A+2g%7D+%3D++%5Csqrt%7B2gh%7D+%3D++%5Csqrt%7B2%2A10%2A25%7D+%3D+++%5Csqrt%7B500)
≈ 22,36 м/c
S=29*10^-2*0,5*10^-3=14,5*10^-5 м2
P=F/S=600/14,5*10^-5=41,4*10^5 H