This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Kutokana na mchoro, bomba linaonyeshwa kama silinda yenye mashimo, lakini kipenyo cha ndani au unene wa ukuta haujatolewa. Kwa hivyo, tutachukulia bomba kama silinda imara (solid cylinder) kwa mahesabu haya, tukitumia vipimo vilivyotolewa. Vipimo vilivyotolewa: Urefu wa bomba (h) = 2.5 M Kipenyo cha nje (D) = 25 cm Step 1: Badilisha vipimo vyote kuwa mita. Urefu h = 2.5 m Kipenyo D = 25 cm = (25)/(100) m = 0.25 m Kisha, kokotoa rediasi (r): r = (D)/(2) = 0.25 m2 = 0.125 m Step 2: Kokotoa Eneo la Jumla la Uso (Total Surface Area) la silinda. Fomula ya eneo la jumla la uso wa silinda imara ni A = 2 r h + 2 r^2. A = 2 (0.125 m)(2.5 m) + 2 (0.125 m)^2 A = 2 (0.3125 m^2) + 2 (0.015625 m^2) A = 0.625 m^2 + 0.03125 m^2 A = (0.625 + 0.03125) m^2 A = 0.65625 m^2 Kutumia ≈ 3.14159: A ≈ 0.65625 × 3.14159 m^2 A ≈ 2.0617 m^2 Eneo la Jumla la Uso ni: 2.0617 m^2 Step 3: Kokotoa Kiasi (Volume) cha shaba kwenye bomba. Fomula ya kiasi cha silinda imara ni V = r^2 h. V = (0.125 m)^2 (2.5 m) V = (0.015625 m^2) (2.5 m) V = 0.0390625 m^3 Kutumia ≈ 3.14159: V ≈ 0.0390625 × 3.14159 m^3 V ≈ 0.1227 m^3 Kiasi cha shaba kwenye bomba ni: 0.1227 m^3 Tuma swali linalofuata 📸