Физика
Contents
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Физика#
Богданов Алексей Александрович
Каждый 8 недель тест, контрольная
Разбаловка#
40 за контрольные (60% +)
10 за активность на семах, явка
50 за экзамен
Литература#
Задачник: занятие 1, занятие 2, занятие 3, занятие 4, занятие 5
Полезная база: 10 класс, 11 класс
Разборы того, что проходим тут
СЕМ 1#
Материальная точка в пространстве обозначается радиус-вектором \(\vec{r}\)
\(\vec{r}(t) = x\vec{e_x} + y\vec{e_y} + z\vec{e_z}\)
\(\vec{v}(t) = \vec{r}_t'(t)=x'(t)\vec{e_x}+y'(t)\vec{e_y} + z'(t)\vec{e_z}\)
\(\vec{w}(t) = \vec{a} = \vec{v_t}'(t) = \vec{\Gamma}_{tt}''(t)\)
\(\vec{r}(t) = \vec{e_x}2t^2 + \vec{e_y}t^2 + \vec{e_z}\) \(\vec{r}'(t) = 4t\vec{e_x} + 2t\vec{e_y}\) \(|\vec{r}(t)|^2 = \vec{r}(t)\vec{r}(t) = \sqrt{4t^4 + t^2 + 1} = \sqrt{5t^2 + 1}\)
\(\vec{r}_1 = \vec{e}_x +3\vec{e}_y + 5\vec{e_z}, \vec{r}_2 = 2\vec{e}_x +4\vec{e}_y + 5\vec{e_z}\) \(\Delta\vec{r} = \vec{r}-2-\vec{r}_1 = \vec{e_x} + \vec{e_y} + \vec{e_z}\) \(|\Delta\vec{r}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\)
\(\vec{r}(t) = \vec{e_x}3t^2 + \vec{e_y}2t + \vec{e_z}\) \(\vec{v}(t) = \vec{r}'(t) = 6t\vec{e_x} + 2\vec{e_y}\) \(\vec{w}(t) = \vec{v}'(t) = 6\vec{e_x}\) \(\vec{v}(1) = 6\vec{e_x} + 2\vec{e}_y = \sqrt{6^2+2^2}=\sqrt{40}\)
\(\vec{r}(t) = \vec{a}t(1-\alpha t)\) \(\vec{r}(t) = \vec{a}t - \vec{a}\alpha t^2\) \(\vec{v}(t)=\vec{r}'(t)=\vec{a}-2\vec{a}\alpha t = \vec{a}(1 - 2\alpha t)\) \(\vec{w}(t) = \vec{v}'(t) = -2\vec{a}\alpha\) \(\Delta t - ?\) \(\vec{S}=\vec{v}_at + \frac{\vec{w}t^2}{2}\) \(\vec{v_0}=\vec{a}\) \(t_0=0\) \(\vec{a}t -\frac{a\vec{a}\alpha t^2}{2}=0\) \(\vec{a}t - \vec{a}\alpha t^2=0\) \(\vec{a}t(1-\alpha t) = 0\) \(t=0 or \alpha t=1\) \(t_{left} = \frac{1}{2\alpha}\) \(S = \frac{4\vec{a}}{4\alpha} - \frac{2\vec{a}}{4\alpha} = \frac{\vec{a}}{2\alpha}\)
\(v=\alpha \sqrt{x}, a>0\) \(t=0\ and\ x=0\) \(t = \_v\ \_a -\ ?\) \(\vec{v}_{mid}\ -\ ?\) \(\int{\frac{dx}{\sqrt{x}}}=\int d\times dt\) \(\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2\sqrt{x} = dt + const\) \(x(0)=O=2\sqrt{O}=dO+const\) \(const = 0\) \(\braket{v}=\frac{S}{t}\) \(S = \frac{\alpha^2t^2}{4} \rightarrow t=\frac{2}{\alpha}\sqrt{5}\)
\(\vec{v}=\vec{r}'(t) = \frac{d\vec{r}}{dt}\)
\(v_c=\frac{dx}{dt}=\alpha \sqrt{x}\)
\(\frac{3}{4}S - v=60km/h\) \(\frac{1}{2}S - v=80km/h\) \(v_{mid}=S/t=\frac{S}{t_1+t_2}=\frac{S}{(3/4)(S/60)+(1/4)(s/80)} = \frac{320}{5} = 64\)
\(w=\frac{dv}{dt} = \alpha\sqrt{v}, \alpha>0\) \(\int\frac{1}{\sqrt{v}}dv=\int{\alpha dt}\) \(2\sqrt{v}+\alpha t + const \rightarrow const = \sqrt{v_0}\) \(t=0 \rightarrow w\sqrt{v} = -\alpha t \uparrow2\) \(4v = \alpha^2-4\alpha\sqrt{2v}t + 4v_0\) \(v=v_0 - \alpha \sqrt{v_0}t + \frac{a^2t^2}{4}\) \(v_0-\alpha\sqrt{v}t+ \frac{\alpha t^2}{4} = 0\) \(D = \alpha^2v_0-\alpha^2v_0 = 0\) \((\frac{\alpha}{2}t - \sqrt{v_0}) = 0, t=\frac{2\sqrt{v_0}}{\alpha}\) \(S=\int^{\frac{2\sqrt{v_0}}{\alpha}}_0(\frac{\alpha^2t^2}{4}-\alpha\sqrt{v_0}t + v_0)dt=2 = (\frac{\alpha^2}{4} \frac{3}{3} - \alpha\frac{v_0}{2}t^2 + v_0) |^{\frac{2\sqrt{v_0}}{\alpha}}_0 = \frac{\alpha^2}{12}\frac{8\sqrt{v_0^3}}{a^3} - \alpha\frac{v_0}{2}\frac{4v_0}{\alpha^2} + \frac{2\sqrt{v_0^3}}{\alpha} = \frac{2v_0\sqrt{v_0}}{3\alpha}\)
ДЗ: 10, 11, 12, 15, 13 (проверяется)
СаСавельев, механика, 1 том
\(h = v_0 + \frac {gt^2} 2\)
\(10 = 15t + \frac {10t^2} 2\)
\(t_1\)
\(\tau = t_1\frac {10} {340}\)
\(x(t) = x_0 +v_0t + \frac{at^2}2\)
ДЗ: 1, 2, 3, 4, 10, 13, 14, 15
№1
Дано
\(t_0 = ol\)
\(a) V-?\)
b) \(W\)
\(\vec a = -\vec R\)
а) \(V = \frac{RV_0}{R+V_0t}\)
\(W_1 = \frac {V^2} R = \frac {dV}{dt} = \int \frac {dv}{r^2}\thus\int \frac {dt}{R}\)
\(\frac {-1} V + c = \frac t R\)
\(V(0) = V_0\)
\(\frac 1 V = \frac{R-V_0t}{V_0R}\)
“Куда хотите, туда и суйте”
\(-\frac 2 {V_0} +c = 0\)
\(c = \frac 2 {V_0}\)
\(\frac t R = \frac 1 {V_0} - \frac 2 V\)
\(\frac 1 V = \frac 1 {V_0} - \frac t R\)
\(V(t) = \frac {dS}{st}\)
\(\frac {V_0R}{R+V_0t}=\frac{dS}{st}\)
\(\int_0^t \frac{dtV_0R}{R+V_0t} =\int_0^tdS\)
\(R\int_0^t \frac {d(V_0t+R)}{R+V_0t}=S\)
\(R(ln(R+V_0t) - ln(R)) = S\)
\(R(ln(\frac{R+V_0t}{R})) = S\)
\(\frac S R = ln(\frac{V_0t+R}R)\)
\(e = \frac{V_0t + R}R\)
\(V(S) = V_0\frac1 e S\)
\(V(S) = V_0 \frac R e\)
№2
Дано
\(\phi = at-bt^3\)
\(a = 6\) рад/сек
\(b=2\) рад/сек
Найти
a) Среднее \(\phi'\) и \(\phi''\) при \(t\) от 0 до остановки
b) \(\phi''\) в момент остановки \(\bkets{w}=2a/3, \bkets{\beta}=\sqrt{3ab}, |\beta_{кон}|=2\sqrt{3ab}\)
Решение
a) \(w(t)=\phi'(t)=a-3bt^2\)
\(a(t)=-6bt\)
\(a=3bt^2=0\)
\(t=\sqrt{\frac a {3b}}\)
\(\bkets{w} = \frac{a\sqrt{\frac{a}{3b}} - b\sqrt{\frac{a}{3b}} \frac{a}{3b}}{\sqrt{\frac{2}{3b}}} = \frac{2a}3\)
\(<a> = \sqrt{3ab}\)
\(a(\sqrt{a/3b}) = -6b\sqrt{a/3b} = -\sqrt{12ab}\)
№3
Дано
\(t=2,5\)с
\(a=0,2\)рад/с
\(g=0,65\)м/с
Решение
\(\phi'=2at\)
\(v=wR\)
\(v=2atR\)
\(R=\frac v {2at}\)
\(w_t = \frac {dv}{dt} = 2aR\)
\(w_k = \frac{v^22at}{v}=v2at=\frac{2av}{2at} = \frac v t\)
\(w_k = \sqrt{w_t^2 + w_k^2} = \sqrt{(\frac v t)^2+(v2at)^2}\)
№4
\(\beta = 3\) рад/с
\(R-?\)
при \(t=1\) \(a=7,5м/с^2\)
\(\phi_k=\frac {v^2}R\)
\(\beta =\frac {dw}{dt}\), т.к. \(\beta = const \thus \beta = \frac{w_0}t\)
\(w = \beta t\)
\(v = wR = \beta t R\)
\(\phi_t = \beta R\)
\(a = \sqrt{\phi_k^2+\phi_t^2}=\sqrt{\beta^2R^2 + \frac {S^4}{R^2}}\)
№10
\(v_k=320\)м/с
\(n=2\)
\(l=2\)м
\(a=const\)
\(w-?\)
\(\omega = 2\pi n v / L = 2*10^3\)рад/с
\(v_k = at \thus a=\frac{v_k}t}\)
\(L=\frac{at^2}2 \thus L=\frac{v_k}{2} \thus \t = \frac{2L}{v_k}\)
(не корректно) \(n2\pi=\omega t \thus \omega = 2\pi n v_k / 2e \thus \frac{\pi n v_k}{e}\)
\(2\pi n = \frac{\beta t^2}2\) \(\frac{wn}{2\pi n}=\frac{2}{t}\)
\(w_k = \beta t\)
\(w_k = \frac{4\pi n}t = \frac{4\pi n}{2e} v_k = \frac{2\pi n}e v_k\)
Тест на пятой недел (25-30 мин)
СЕМ 20/09/22#
Савельев, механика, 1 том
\(h = v_0 + \frac {gt^2} 2\)
\(10 = 15t + \frac {10t^2} 2\)
\(t_1\)
\(\tau = t_1\frac {10} {340}\)
\(x(t) = x_0 +v_0t + \frac{at^2}2\)
ДЗ: 1, 2, 3, 4, 10, 13, 14, 15
№1
Дано
\(t_0 = ol\)
\(a) V-?\)
b) \(W\)
\(\vec a = -\vec R\)
а) \(V = \frac{RV_0}{R+V_0t}\)
\(W_1 = \frac {V^2} R = \frac {dV}{dt} = \int \frac {dv}{r^2}\thus\int \frac {dt}{R}\)
\(\frac {-1} V + c = \frac t R\)
\(V(0) = V_0\)
\(\frac 1 V = \frac{R-V_0t}{V_0R}\)
“Куда хотите, туда и суйте”
\(-\frac 2 {V_0} +c = 0\)
\(c = \frac 2 {V_0}\)
\(\frac t R = \frac 1 {V_0} - \frac 2 V\)
\(\frac 1 V = \frac 1 {V_0} - \frac t R\)
\(V(t) = \frac {dS}{st}\)
\(\frac {V_0R}{R+V_0t}=\frac{dS}{st}\)
\(\int_0^t \frac{dtV_0R}{R+V_0t} =\int_0^tdS\)
\(R\int_0^t \frac {d(V_0t+R)}{R+V_0t}=S\)
\(R(ln(R+V_0t) - ln(R)) = S\)
\(R(ln(\frac{R+V_0t}{R})) = S\)
\(\frac S R = ln(\frac{V_0t+R}R)\)
\(e = \frac{V_0t + R}R\)
\(V(S) = V_0\frac1 e S\)
\(V(S) = V_0 \frac R e\)
№2
Дано
\(\phi = at-bt^3\)
\(a = 6\) рад/сек
\(b=2\) рад/сек
Найти
a) Среднее \(\phi'\) и \(\phi''\) при \(t\) от 0 до остановки
b) \(\phi''\) в момент остановки \(\bkets{w}=2a/3, \bkets{\beta}=\sqrt{3ab}, |\beta_{кон}|=2\sqrt{3ab}\)
Решение
a) \(w(t)=\phi'(t)=a-3bt^2\)
\(a(t)=-6bt\)
\(a=3bt^2=0\)
\(t=\sqrt{\frac a {3b}}\)
\(\bkets{w} = \frac{a\sqrt{\frac{a}{3b}} - b\sqrt{\frac{a}{3b}} \frac{a}{3b}}{\sqrt{\frac{2}{3b}}} = \frac{2a}3\)
\(<a> = \sqrt{3ab}\)
\(a(\sqrt{a/3b}) = -6b\sqrt{a/3b} = -\sqrt{12ab}\)
№3
Дано
\(t=2,5\)с
\(a=0,2\)рад/с
\(g=0,65\)м/с
Решение
\(\phi'=2at\)
\(v=wR\)
\(v=2atR\)
\(R=\frac v {2at}\)
\(w_t = \frac {dv}{dt} = 2aR\)
\(w_k = \frac{v^22at}{v}=v2at=\frac{2av}{2at} = \frac v t\)
\(w_k = \sqrt{w_t^2 + w_k^2} = \sqrt{(\frac v t)^2+(v2at)^2}\)
№4
\(\beta = 3\) рад/с
\(R-?\)
при \(t=1\) \(a=7,5м/с^2\)
\(\phi_k=\frac {v^2}R\)
\(\beta =\frac {dw}{dt}\), т.к. \(\beta = const \thus \beta = \frac{w_0}t\)
\(w = \beta t\)
\(v = wR = \beta t R\)
\(\phi_t = \beta R\)
\(a = \sqrt{\phi_k^2+\phi_t^2}=\sqrt{\beta^2R^2 + \frac {S^4}{R^2}}\)
№10
\(v_k=320\)м/с
\(n=2\)
\(l=2\)м
\(a=const\)
\(w-?\)
\(\omega = 2\pi n v / L = 2*10^3\)рад/с
\(v_k = at \thus a=\frac{v_k}t}\)
\(L=\frac{at^2}2 \thus L=\frac{v_k}{2} \thus \t = \frac{2L}{v_k}\)
(не корректно) \(n2\pi=\omega t \thus \omega = 2\pi n v_k / 2e \thus \frac{\pi n v_k}{e}\)
\(2\pi n = \frac{\beta t^2}2\) \(\frac{wn}{2\pi n}=\frac{2}{t}\)
\(w_k = \beta t\)
\(w_k = \frac{4\pi n}t = \frac{4\pi n}{2e} v_k = \frac{2\pi n}e v_k\)
Тест на пятой недел (25-30 мин)
Савельев, механика, 1 том
\(h = v_0 + \frac {gt^2} 2\)
\(10 = 15t + \frac {10t^2} 2\)
\(t_1\)
\(\tau = t_1\frac {10} {340}\)
\(x(t) = x_0 +v_0t + \frac{at^2}2\)
ДЗ: 1, 2, 3, 4, 10, 13, 14, 15
№1
Дано
\(t_0 = ol\)
\(a) V-?\)
b) \(W\)
\(\vec a = -\vec R\)
а) \(V = \frac{RV_0}{R+V_0t}\)
\(W_1 = \frac {V^2} R = \frac {dV}{dt} = \int \frac {dv}{r^2}\thus\int \frac {dt}{R}\)
\(\frac {-1} V + c = \frac t R\)
\(V(0) = V_0\)
\(\frac 1 V = \frac{R-V_0t}{V_0R}\)
“Куда хотите, туда и суйте”
\(-\frac 2 {V_0} +c = 0\)
\(c = \frac 2 {V_0}\)
\(\frac t R = \frac 1 {V_0} - \frac 2 V\)
\(\frac 1 V = \frac 1 {V_0} - \frac t R\)
\(V(t) = \frac {dS}{st}\)
\(\frac {V_0R}{R+V_0t}=\frac{dS}{st}\)
\(\int_0^t \frac{dtV_0R}{R+V_0t} =\int_0^tdS\)
\(R\int_0^t \frac {d(V_0t+R)}{R+V_0t}=S\)
\(R(ln(R+V_0t) - ln(R)) = S\)
\(R(ln(\frac{R+V_0t}{R})) = S\)
\(\frac S R = ln(\frac{V_0t+R}R)\)
\(e = \frac{V_0t + R}R\)
\(V(S) = V_0\frac1 e S\)
\(V(S) = V_0 \frac R e\)
№2
Дано
\(\phi = at-bt^3\)
\(a = 6\) рад/сек
\(b=2\) рад/сек
Найти
a) Среднее \(\phi'\) и \(\phi''\) при \(t\) от 0 до остановки
b) \(\phi''\) в момент остановки \(\bkets{w}=2a/3, \bkets{\beta}=\sqrt{3ab}, |\beta_{кон}|=2\sqrt{3ab}\)
Решение
a) \(w(t)=\phi'(t)=a-3bt^2\)
\(a(t)=-6bt\)
\(a=3bt^2=0\)
\(t=\sqrt{\frac a {3b}}\)
\(\bkets{w} = \frac{a\sqrt{\frac{a}{3b}} - b\sqrt{\frac{a}{3b}} \frac{a}{3b}}{\sqrt{\frac{2}{3b}}} = \frac{2a}3\)
\(<a> = \sqrt{3ab}\)
\(a(\sqrt{a/3b}) = -6b\sqrt{a/3b} = -\sqrt{12ab}\)
№3
Дано
\(t=2,5\)с
\(a=0,2\)рад/с
\(g=0,65\)м/с
Решение
\(\phi'=2at\)
\(v=wR\)
\(v=2atR\)
\(R=\frac v {2at}\)
\(w_t = \frac {dv}{dt} = 2aR\)
\(w_k = \frac{v^22at}{v}=v2at=\frac{2av}{2at} = \frac v t\)
\(w_k = \sqrt{w_t^2 + w_k^2} = \sqrt{(\frac v t)^2+(v2at)^2}\)
№4
\(\beta = 3\) рад/с
\(R-?\)
при \(t=1\) \(a=7,5м/с^2\)
\(\phi_k=\frac {v^2}R\)
\(\beta =\frac {dw}{dt}\), т.к. \(\beta = const \thus \beta = \frac{w_0}t\)
\(w = \beta t\)
\(v = wR = \beta t R\)
\(\phi_t = \beta R\)
\(a = \sqrt{\phi_k^2+\phi_t^2}=\sqrt{\beta^2R^2 + \frac {S^4}{R^2}}\)
№10
\(v_k=320\)м/с
\(n=2\)
\(l=2\)м
\(a=const\)
\(w-?\)
\(\omega = 2\pi n v / L = 2*10^3\)рад/с
\(v_k = at \thus a=\frac{v_k}t}\)
\(L=\frac{at^2}2 \thus L=\frac{v_k}{2} \thus \t = \frac{2L}{v_k}\)
(не корректно) \(n2\pi=\omega t \thus \omega = 2\pi n v_k / 2e \thus \frac{\pi n v_k}{e}\)
\(2\pi n = \frac{\beta t^2}2\) \(\frac{wn}{2\pi n}=\frac{2}{t}\)
\(w_k = \beta t\)
\(w_k = \frac{4\pi n}t = \frac{4\pi n}{2e} v_k = \frac{2\pi n}e v_k\)
Тест на пятой недел (25-30 мин)
ЛЕК 22/09/22#
(сами догадайтесь, что он шептал)
Относительное движение систем отсчёта
\(\cases{\vec r (t) = \vec r'(t) \vec V_0 t \\ t = t'}\)
\(\frac{d\vec r}{dt} = \frac{d\vec r}{dt} \thus \vec V = \vec V' +\vec {V_0}\)
\(\frac {d\vec V}{dt} = \frac{d\vec V'}{dt}\thus \vec W = \vec W'\)
\(\cases{x(t)=x'(t)+v_0t \\ y(t)=y'(t) \\ z(t) = z'(t) \\ t'=t}\)
Всегда существуют (тангенцияальные) системы отсчёта, в котрых тело покоится, или двигается линейно
Мера инертности тела
\(m=\sum_nm_i\)
Зависимость массы тела
\(\frac {m_1}{m_2} = \frac {W_2}{W_1}\)
\(\vec F = m\vec w\)
\(\vec F = \sum_n \vec {F_i}\)
\(m\vec w = n\frac{d\vec V}{dt}=\frac{d(m\vec V)}{dt}=\frac{d\vec p}{dt}\) \(\vec p \up{def}= m\vec V \thus \vec F = \frac{d\vec p}{dt}\)
\(\vec F_{22} = - \vec F_{21}\)
Решение задач по динамике:
изобразить силы, действующие на тело
изобразить оси
записать основное уравнени динамики в векторной форме
Решаем систему уравнений \(\cases{Ox: m\vec g \sin \alp = 0 \\ Oy: m\vec g \cos \alp = 0}\)
Сила сопротивления среды#
\(\vec F_{сопр} = -k(v)\vec e_vV^n,\ n=1,2\ \vec e_V=\frac{\vec V}V\)
\(\vec F_{сопр} = -k(v)\vec e_v, \vec V \leq V_{гр}\)
\(\vec F_{сопр} = -k(v)\vec e_vV^2, \vec V > V_{гр}\)
\(F_{тр} = k F_{тр}\)
\(F_{тр} = k N\)
Упругость#
\(\vec F_{упр} = -\vec F_{вн} \thus F_{упр} = F_{вн} = k|x|\)
\(\vec F_{упр} = -kx\)
\(\Delta l \geq 0\)
Деформация#
\(\eps = \frac {\Delta l}{l_0}\)
При упругой дефорации пропорционально силе, приходейся на площадь поперечного сечениея
\(\eps \proportional \frac F S\)
\(\eps = \alp \frac F S\)
Велиина, равная отношению силы, действующая на поверхности этой силы? называется напряжением
Если сила F направлена по нормали поверхности S, то напряжение называется нормалью
Если же сила F направлена по касательной поверхности, на короую действует, то и напряжение называется тангенсальным
Нормальное напряжение \(\sigma\)
\(\eps = \alp \sigma = \frac \sigma E\)
\(\frac 1 \alp E\)
\([E] = \frac H {M^2}\)
\(F = \eps \)
СЕМ 04.10.22#
№ 2
\(\frac{T_{юс}}{T_{зс}} = 12\)
\(\frac{e_ю}{e_з} - ?\)
Для Юпитера:
\(M_юa=G\frac{M_юM_с}{l^2_ю} | \frac{e^2_ю}{M_ю}\)
\(a = \frac{4\pi^2R}{T^2_ю}\)
\(\frac{e^2_ю4\pi^2R}{T^2_ю} = Gm_с\)
Для земли:
\(M_з4\pi l_з = G\frac{M_зM_с}{e^2_з} | \frac{l^2_з}{M_з}\)
\(GM_c=\frac{4\pi^2l^3_з}{T^2_з} = \frac{4\pi^2l^3_ю}{T^2_ю} | \frac{1}{4\pi^2}\)
\(\frac{l_ю}{l_з} = root 3 {{\frac{T_ю}{T_з}}^2}\)
\(a_ю = \frac{F}{M_ю} = \frac{GM_юM_c}{l^2_юM_ю}\)
\(a = \frac{GM_c}{l^2_ю}\)
\(a=\frac{v^2}{l_ю}\)
\(v = \sqrt{al_ю} = \sqrt{\frac{GM_c}{l_ю}}\)
\(\ro=const\)
\(3M_1 = M_2\)
\(F_1 = G\frac{M_1M_1}{(2R_1)^2}\)
\(F_2 = G\frac{M_2M_2}{(2R_2)^2}\)
\(\frac 4 3 \pi R_1^3\ro = M_1\)
\(\frac 4 3 \pi R_2^3\ro = M_2\)
\(\thus \frac{R_1^3}{R_2^3} = \frac{M_2}{3M_1} = \frac 1 3\)
\(\thus R_2 = root 3 {3R_1^3}\)
\(\frac{F_2}{F_1} = \frac{GM_2^2 4 R_1^2}{GM_1^2 4 R_2^2} = 9 \frac{R_1^2}{R_2^2} = \frac{9}{root 3 {3^2}} = 9^{\frac 2 3} = 3 ^ {\frac 4 3}\)
Дано
\(p = 2.7 г,см^3\)
\(r_1 = 3 см\)
\(r_2 = 5 см\)
\(F=G\frac{m_1m_2}{r_1+r_2}\)
где
\(m_1 = p V_1 = p\frac 4 3 \pi r^3_1\)
\(m_2 = p V_2 = p\frac 4 3 \pi r^3_2\)
\(F = \frac {16} 9 p^2 G \pi^2 \frac{r_1^3r_2^3}{(r_1+r_2)^2}\)
Дано
\(ma=G\frac{m_cM_p}{R^2}\)
\(M_p = p\frac 4 3 \pi R^3\)
\(w^2R=G\frac{p4\pi R^3}{3R^2} = G\frac{p4\pi}{3}\)
\(\frac{4\pi^2}{T^2} = G\frac{p4\pi}{3}\)
\(T = \sqrt{\frac{3\pi}{Gp}}\)
Дано
\(R=30\)
\(T_з=12\)
\(T_н=30T_з\)
\(T_н\)
\(\frac{T_1^2}{T_2^2}=\frac{Q_1^3}{Q_2^3} = 30^3\)
\(T_н=\sqrt{30^3}=\sqrt{27000}\)
СЕМ 21.02.23#
\(Q = \Delta U = A\)
\(A = \int^{V_2}_{V_1} P dv\)
\(U = N \frac 1 2 K T\)
\(U = \frac 1 2 \frac m M R T = \frac i 2 P V\)
\(C_V = \frac i 2 R\)
\(C_P = \frac{i+2} 2 R\)
\(PV^\gamma = const\)