H = 100 м Решение: u = √(2*10*100) ≈ 44,72 м/с
g ≈ 10 Н/кг u = √(2*g*h) t = √((2*100)/10) ≈ 4,47 c
u - ? t = √((2*h)/g)
t - ?
Задача на закон Кулона
заряды берем по модулю
![F=k \frac{q1q2}{R ^{2} }](https://tex.z-dn.net/?f=F%3Dk+%5Cfrac%7Bq1q2%7D%7BR+%5E%7B2%7D+%7D+)
было
![F=k \frac{8q4q}{R ^{2} } =32k \frac{q ^{2} }{R ^{2} }=0.2](https://tex.z-dn.net/?f=F%3Dk+%5Cfrac%7B8q4q%7D%7BR+%5E%7B2%7D+%7D+%3D32k++%5Cfrac%7Bq+%5E%7B2%7D+%7D%7BR+%5E%7B2%7D+%7D%3D0.2+)
<u />
![k \frac{q ^{2} }{R ^{2} }= \frac{0.2}{32} = \frac{1}{160}](https://tex.z-dn.net/?f=k+%5Cfrac%7Bq+%5E%7B2%7D+%7D%7BR+%5E%7B2%7D+%7D%3D+%5Cfrac%7B0.2%7D%7B32%7D+%3D+%5Cfrac%7B1%7D%7B160%7D+)
заряды привели в соприкосновение и развели, заряд каждого
![Q1=Q2=(q1+q2)/2=(-8q+4q)/2=-2q](https://tex.z-dn.net/?f=Q1%3DQ2%3D%28q1%2Bq2%29%2F2%3D%28-8q%2B4q%29%2F2%3D-2q)
стало
![F2=k \frac{Q1Q2}{R ^{2} } =k \frac{2q2q}{R ^{2} } =4*k \frac{q ^{2} }{R ^{2} }=4* \frac{1}{160} = \frac{1}{40} =0.025H](https://tex.z-dn.net/?f=F2%3Dk+%5Cfrac%7BQ1Q2%7D%7BR+%5E%7B2%7D+%7D+%3Dk+%5Cfrac%7B2q2q%7D%7BR+%5E%7B2%7D+%7D+%3D4%2Ak+%5Cfrac%7Bq+%5E%7B2%7D+%7D%7BR+%5E%7B2%7D+%7D%3D4%2A+%5Cfrac%7B1%7D%7B160%7D+%3D+%5Cfrac%7B1%7D%7B40%7D+%3D0.025H)
Ответ
Смотри вложения (там все подробно расписанно) !!!!!!!
<h2><u>Дано</u>:</h2>
Ускорение:
м/с².
Путь:
м.
Соотношение скоростей: ![\bf V = 4V_0.](https://tex.z-dn.net/?f=%5Cbf+V+%3D+4V_0.)
<u>Найти</u> нужно начальную скорость: ![\bf V_0 - ?.](https://tex.z-dn.net/?f=%5Cbf+V_0+-+%3F.)
<h2><u>
Решение</u>:</h2>
1. Зависимость скорости от времени при равноускоренном движении: ![\boxed{\;V = V_0 + at\;}](https://tex.z-dn.net/?f=%5Cboxed%7B%5C%3BV+%3D+V_0+%2B+at%5C%3B%7D)
2. Путь при равноускоренном движении: ![\boxed{\;S = V_0t + \dfrac{at^2}{2}\;}](https://tex.z-dn.net/?f=%5Cboxed%7B%5C%3BS+%3D+V_0t+%2B+%5Cdfrac%7Bat%5E2%7D%7B2%7D%5C%3B%7D)
<h3>Решаем систему из (1) и (2).</h3>
![\begin{cases}4V_0 = V_0 + 5t,\\600 = V_0t + \dfrac{5t^2}{2};\end{cases} \Longleftrightarrow\;\;\begin{cases}t = \dfrac{3}{5}V_0,\\600 = V_0t + \dfrac{5t^2}{2};\end{cases} \Longrightarrow\\\\600 = \dfrac{3}{5}V_0^2 + \dfrac{5}{2}\cdot\dfrac{9}{25}V_0^2;\\600 = V_0^2\bigg(\dfrac{3}{5} + \dfrac{9}{10}\bigg);\\600 = 1,5\cdot V_0^2\\V_0^2 = \dfrac{600}{1,5};\\V_0^2 = 400;\\](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D4V_0+%3D+V_0+%2B+5t%2C%5C%5C600+%3D+V_0t+%2B+%5Cdfrac%7B5t%5E2%7D%7B2%7D%3B%5Cend%7Bcases%7D+%5CLongleftrightarrow%5C%3B%5C%3B%5Cbegin%7Bcases%7Dt+%3D+%5Cdfrac%7B3%7D%7B5%7DV_0%2C%5C%5C600+%3D+V_0t+%2B+%5Cdfrac%7B5t%5E2%7D%7B2%7D%3B%5Cend%7Bcases%7D+%5CLongrightarrow%5C%5C%5C%5C600+%3D+%5Cdfrac%7B3%7D%7B5%7DV_0%5E2+%2B+%5Cdfrac%7B5%7D%7B2%7D%5Ccdot%5Cdfrac%7B9%7D%7B25%7DV_0%5E2%3B%5C%5C600+%3D+V_0%5E2%5Cbigg%28%5Cdfrac%7B3%7D%7B5%7D+%2B+%5Cdfrac%7B9%7D%7B10%7D%5Cbigg%29%3B%5C%5C600+%3D+1%2C5%5Ccdot+V_0%5E2%5C%5CV_0%5E2+%3D+%5Cdfrac%7B600%7D%7B1%2C5%7D%3B%5C%5CV_0%5E2+%3D+400%3B%5C%5C)
(м/с).
<h2><u>
Ответ</u>: 20 м/с.</h2>