This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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10. Kuamua urefu wa kunyooshwa (extension) unaotokana na chemchemi mbili zinazofanana zilizounganishwa sambamba wakati uzito wa 0.2 \, kg umening'inizwa chini yake. Step 1: Tafuta kani ya jumla ya chemchemi (effective spring constant) kwa chemchemi zilizounganishwa sambamba. Kani ya chemchemi moja k = 1000 \, Nm^-1. Kwa chemchemi mbili sambamba, kani ya jumla k_jumla = k_1 + k_2. k_jumla = 1000 \, Nm^-1 + 1000 \, Nm^-1 = 2000 \, Nm^-1 Step 2: Kokotoa nguvu inayotokana na uzito. Nguvu F = mg, ambapo m = 0.2 \, kg na g = 9.8 \, ms^-2. F = 0.2 \, kg × 9.8 \, ms^-2 = 1.96 \, N Step 3: Tumia Sheria ya Hooke kupata urefu wa kunyooshwa x. Sheria ya Hooke inasema F = k_jumlax. x = (F)/(k_jumla) = 1.96 \, N2000 \, Nm^-1 = 0.00098 \, m Urefu wa kunyooshwa ni: 0.00098 \, m 11. Mchoro unaonyesha chupa ya koni yenye maji hadi kiwango kilichoonyeshwa. a) Eleza mabadiliko katika utulivu wa chupa wakati maji zaidi yameongezwa ndani yake. Utulivu wa chupa utaongezeka. b) Toa sababu ya jibu lako katika a) hapo juu. Kuongeza maji kunapunguza kituo cha mvuto (center of gravity) cha chupa na maji, na kituo cha mvuto cha chini huongeza utulivu. 12. a) Gari lilianza kusonga kwa kasi ya 10 \, ms^-1 na kuongeza kasi kwa usawa kwa 1 \, ms^-2 hadi kufikia kasi ya 15 \, ms^-1. Kokotoa: i) Muda uliotumika. Tunatumia fomula ya mwendo: v = u + at. Hapa, u = 10 \, ms^-1, v = 15 \, ms^-1, na a = 1 \, ms^-2. 15 \, ms^-1 = 10 \, ms^-1 + (1 \, ms^-2)t 15 - 10 = t t = 5 \, s Muda uliotumika ni: 5 \, s ii) Umbali uliosafirishwa wakati wa kuongeza kasi. Tunatumia fomula ya mwendo: s = ut + (1)/(2)at^2. Hapa, u = 10 \, ms^-1, t = 5 \, s, na a = 1 \, ms^-2. s = (10 \, ms^-1)(5 \, s) + (1)/(2)(1 \, ms^-2)(5 \, s)^2 s = 50 \, m + (1)/(2)(1 \, ms^-2)(25 \, s^2) s = 50 \, m + 12.5 \, m s = 62.5 \, m Umbali uliosafirishwa ni: 62.5 \, m iii) Kasi iliyofikiwa mita 100 kutoka mahali ambapo kuongeza kasi kulianza. Tunatumia fomula ya mwendo: v^2 = u^2 + 2as. Hapa, u = 10 \, ms^-1, a = 1 \, ms^-2, na s = 100 \, m. v^2 = (10 \, ms^-1)^2 + 2(1 \, ms^-2)(100 \, m) v^2 = 100 \, m^2s^-2 + 200 \, m^2s^-2 v^2 = 300 \, m^2s^-2 v = sqrt(300) \, ms^-1 v ≈ 17.32 \, ms^-1 Kasi iliyofikiwa ni: 17.32 \, ms^-1 Tuma swali linalofuata 📸