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Forward speed: 5 km/hr Field efficiency: 80% Row spacing: 200 mm Forward speed = 5 km/hr = 5,000 meters / 60 minutes = 83.33 meters/min Working width = 9 rows x 200 mm = 1,800 mm = 1.8 meters Effective field capacity = (Working width x Forward speed x 10) / Field efficiency = (1.8 meters x 83.33 meters/min x 10) / 0.80 = 2249.92 square meters per minute To convert the field capacity from square meters per minute to hectares per hour: Effective field capacity = 2249.92 square meters/minute x (60 minutes/10,000 square meters) Effective field capacity = 1.35 hectares per hour Therefore, the effective field capacity of the 9-row seed cum fertilizer drill operated at a forward speed of 5 km/hr with a field efficiency of 80% and a row spacing of 200 mm is approximately 1.35 hectares per hour.
I. x2 – 7x + 12 = 0
II. y2 – 7y + 10 = 0
Let 's' represent the sum of the highest root of equations I and III, and 'r' denote the product of the lowest root of equation I and the highest root o...
(i) 2x² + 14x - 16 = 0
(ii) y² – y – 12 = 0
I. 22x² - 97x + 105 = 0
II. 35y² - 61y + 24 = 0
LCM of 'x' and 'y' is 30 and their HCF is 1 such that {10 > x > y > 1}.
I. 2p²- (x + y) p + 3y = 0
II. 2q² + (9x + 2) = (3x + y) q
I. 2x² - 15x + 27 = 0
II. 2y² - 13y + 20 = 0
I. 5q = 7p + 21
II. 11q + 4p + 109 = 0
One of the roots of the equation x² – 12x + k = 0 is x = 3. The other root is ___________.
I. 9x2 + 45x + 26 = 0
II. 7y2 – 59y − 36 = 0