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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Morning Water — let's get this done. Hapa kuna hesabu ya vipingamizi vinavyofaa kwa kila mchoro: 1. Mchoro wa 1: Step 1: Tambua vipingamizi vilivyounganishwa kwa mfululizo katika tawi la chini la sambamba. R_chini = 1\, + 2\, + 1\, = 4\, Step 2: Kokotoa kipingamizi kinachofaa cha vipingamizi vilivyounganishwa kwa sambamba (tawi la juu la 4\, na tawi la chini la 4\,). R_sambamba = (4\, × 4\,)/(4\, + 4\,) = (16\,^2)/(8\,) = 2\, Step 3: Kokotoa kipingamizi kinachofaa cha jumla kwa kuongeza kipingamizi cha mwanzo cha 2\, na kipingamizi cha sambamba. R_jumla = 2\, + 2\, = 4\, Kipingamizi kinachofaa kwa mchoro wa 1 ni: 4\, 2. Mchoro wa 2: Step 1: Kokotoa kipingamizi cha jumla cha tawi la juu la sambamba. R_juu = 50\, + 200\, = 250\, Step 2: Kokotoa kipingamizi cha jumla cha tawi la chini la sambamba. R_chini = 300\, + 100\, = 400\, Step 3: Kokotoa kipingamizi kinachofaa cha vipingamizi vilivyounganishwa kwa sambamba. R_sambamba = (250\, × 400\,)/(250\, + 400\,) = (100000\,^2)/(650\,) = (10000)/(65)\, = (2000)/(13)\, Step 4: Kokotoa kipingamizi kinachofaa cha jumla kwa kuongeza kipingamizi cha sambamba na kipingamizi cha mwisho cha 250\,. R_jumla = (2000)/(13)\, + 250\, = (2000 + (250 × 13))/(13)\, = (2000 + 3250)/(13)\, = (5250)/(13)\, Kipingamizi kinachofaa kwa mchoro wa 2 ni: (5250)/(13)\, ≈ 403.85\, 3. Mchoro wa 3: Step 1: Kokotoa kipingamizi cha jumla cha tawi la chini la sambamba. R_chini = 3\, + 4\, = 7\, Step 2: Kokotoa kipingamizi kinachofaa cha vipingamizi vilivyounganishwa kwa sambamba (tawi la juu la 6\, na tawi la chini la 7\,). R_sambamba = (6\, × 7\,)/(6\, + 7\,) = (42\,^2)/(13\,) = (42)/(13)\, Step 3: Kokotoa kipingamizi kinachofaa cha jumla kwa kuongeza vipingamizi vyote vilivyounganishwa kwa mfululizo. R_jumla = 5\, + (42)/(13)\, + 2\, = 7\, + (42)/(13)\, = ((7 × 13) + 42)/(13)\, = (91 + 42)/(13)\, = (133)/(13)\, Kipingamizi kinachofaa kwa mchoro wa 3 ni: (133)/(13)\, ≈ 10.23\, 4. Mchoro wa 4: Step 1: Kokotoa kipingamizi cha jumla cha tawi la chini la sambamba. R_chini = 4\, + 1\, = 5\, Step 2: Kokotoa kipingamizi kinachofaa cha vipingamizi vilivyounganishwa kwa sambamba (tawi la juu la 1\, na tawi la chini la 5\,). R_sambamba = (1\, × 5\,)/(1\, + 5\,) = (5\,^2)/(6\,) = (5)/(6)\, Step 3: Kokotoa kipingamizi kinachofaa cha jumla kwa kuongeza vipingamizi vyote vilivyounganishwa kwa mfululizo. R_jumla = 2\, + (5)/(6)\, + 2\, = 4\, + (5)/(6)\, = ((4 × 6) + 5)/(6)\, = (24 + 5)/(6)\, = (29)/(6)\, Kipingamizi kinachofaa kwa mchoro wa 4 ni: (29)/(6)\, ≈ 4.83\, Tuma swali linalofuata 📸