Sure, let's break down the solution step by step.
1. Convert the land area into square meters:
- Given land area: \(1.6 \text{ hectares}\)
- \(1 \text{ hectare} = 10,000 \text{ square meters}\)
- Therefore, \(1.6 \text{ hectares} = 1.6 \times 10000 = 16000 \text{ square meters}\)
2. Calculate the effective working width:
- Number of furrows: \(5\)
- Space between furrows: \(10 \text{ cm}\)
- The working width in meters can be calculated by multiplying the number of furrows by the space between them and then converting cm to meters (1 m = 100 cm).
- Effective working width: \(5 \times 10 \text{ cm} = 50 \text{ cm} = 0.5 \text{ meters}\)
3. Convert the speed from km/hr to m/hr:
- Given speed: \(3.2 \text{ km/hr}\)
- \(1 \text{ km} = 1000 \text{ meters}\)
- Therefore, the speed in meters per hour: \(3.2 \times 1000 = 3200 \text{ meters per hour}\)
4. Calculate the area covered per hour without any time loss:
- Area covered per hour is given by the product of the working width and the speed.
- Area covered per hour: \(0.5 \text{ meters} \times 3200 \text{ meters per hour} = 1600 \text{ square meters per hour}\)
5. Consider time loss due to turning (10%):
- Time loss: \(10\%\)
- Actual working hours must account for this time loss. The effective working hours can be found by dividing the total area by the area covered per hour, and then adjusting for time loss.
- Total land area: \(16000 \text{ square meters}\)
- Without time loss, the number of hours needed would be: \(\frac{16000}{1600} = 10 \text{ hours}\)
- Considering the time loss of \(10\%\), we adjust by dividing by \(0.9\) (since \(100\% - 10\% = 90\%\) or \(0.9\)):
- Actual working hours: \(\frac{10}{0.9} \approx 11.111 \text{ hours}\)
Thus, the time required for sowing 1.6 hectares of land by a five-furrow seed drill, taking into account the given factors, is approximately [tex]\(11.11 \text{ hours}\)[/tex].
