Rozwiąż zadania z testu: Proszę o zaznaczenie pogrubionym DANE SZUKANE ROZWIĄZANIE Daje naj za 10 zadań

[latex]Zad.1\ old Dold aold nold e :\Q=100N\F_1=200N\F_2=400N\F_3=100N\g=10 frac{N}{kg} \\old S old z old uold kold aoldnold nold e   :\m = ?\a = ?\\old Rold o old z old w old i old a old z old a old n old i old e :\Q=mg o m = frac{Q}{g} \\a = frac{F}{m} = frac{F_1+F_2-F_3}{m} \\\m = frac{100}{10} =10kg\\a = frac{200+400-100}{10} =50 frac{m}{s^2} [/latex] [latex] Zad.2\old Dold aold nold e :\v_1=0\m_1=20kg\m_2=0,1kg\v_2=800 frac{m}{s}\\ old S old z old uold kold aoldnold nold e:\v_3= ? \\ old Rold o old z old w old i old a old z old a old n old i old e:\p_1+p_2=p_3 o m_1v_1+m_2v_2=(m_1+m_2)cdot v_3\\v_3= frac{m_1v_1+m_2v_2}{m_1+m_2} \\v_3= frac{20cdot 0+0,1cdot800}{20+0,1} approx 3,98 frac{m}{s} [/latex] [latex]Zad.3\old Dold aold nold e:\a=40 frac{m}{s^2}\m=5kg\F_c=240N\\old S old z old uold kold aoldnold nold e :\F_T=?\\old Rold o old z old w old i old a old z old a old n old i old e : \a = frac{F}{m} o F = am\\F_T = F_c - F\\F = 40cdot 5 = 200 N\\F_T =240-200=40 N [/latex] [latex]Zad.4.\old Dold aold nold e :\Delta p = 200 kgcdotfrac{m}{s}\Delta t= 10s\Delta v = 20 frac{m}{s} \\ old S old z old uold kold aoldnold nold e :\F = ?\m=?\a = ?\s = ?\v_s_r = ?\\old Rold o old z old w old i old a old z old a old n old i old e :\p=mcdot v o m = frac{p}{v} \m = frac{200}{20} =10 kg\\a = frac{Delta v }{Delta t} \a = frac{20}{10} =2 frac{m}{s^2} \\a = frac{F}{m} o F = am\F = 2 cdot 10 = 20N\\[/latex] [latex]s = frac{1}{2}at^2\s = frac{1}{2}cdot 2 cdot 10^2=100m\\v_s_r = frac{s}{t} \v_s_r = frac{100}{10}=10 frac{m}{s} [/latex] [latex]Zad.5\old Dold aold nold e :\m=1000kg\s=80m\Delta t = 12s\\old S old z old uold kold aoldnold nold e :\F_o = ?\\old Rold o old z old w old i old a old z old a old n old i old e :\a = frac{F}{m} o F = am\\s = frac{1}{2}at^2\a = frac{2s}{t^2} \\a = frac{2cdot 40}{12^2} =0,5(5) frac{m}{s^2}\\F_o = 0,5(5) cdot 1000 = 555,5(5) N [/latex] [latex]Zad.6\old Dold aold nold e :\Q=100N\a=0,2g = 2 frac{m}{s^2}\\ old S old z old uold kold aoldnold nold e :\m =?\Q_1 = ? \\old Rold o old z old w old i old a old z old a old n old i old e :\a = frac{F}{m } o F = am\Q_1=Q-F\\Q = mg o m= frac{Q}{g} \\\m= frac{100}{10}=10 kg\\F = 2cdot 10 = 20 N \Q_1=100-20=80N[/latex] [latex]Zad.7\old Dold aold nold e :\p = 400 kgcdot frac{m}{s}\v = 36 frac{km}{h} = frac{36000m}{3600s} =10 frac{m}{s} \\ old S old z old uold kold aoldnold nold e :\m = ?\\old Rold o old z old w old i old a old z old a old n old i old e :\p=mv o m = frac{p}{v} \\m = frac{400}{10}=40 kg [/latex] [latex]Zad.8\old Dold aold nold e :\m=10kg\F=200N\\old S old z old uold kold aoldnold nold e :\a =? \\ old Rold o old z old w old i old a old z old a old n old i old e :\a = frac{F}{m} \\a = frac{200}{10} =20 frac{m}{s^2} [/latex] Zad.9 . Druga zasada dynamiki Newtona mówi, że przyspieszenie  z jakim porusza się ciało jest wprost proporcjonalne do siły  działającej na ciało, a odwrotnie proporcjonalne do masy tegoż ciała.  Wzór :  [latex]a = frac{F}{m} \\gdzie :\a - przyspieszenie \F - sila\m - masa[/latex] przykład -> zjeżdżanie z górki rowerem  Zad. 10 Zasada zachowania pedu -> ( W izolowanym układzie ) ciał ich pęd nigdy nie ulega zmianie. : [latex]Delta p_1 = Delta p_2[/latex]