Question 4338Deck General70% to pass
A non-symmetrical tank aboard a MODU contains 390 tons of ballast at a VCG of 9.85 feet. Ballast weighing 250 tons and a VCG of 12.0 feet is discharged. The vertical moments for the remaining ballast is _______________.
AI Explanation
You are correct, the answer A) 842 ft-long tons is the correct answer.
The reasoning is as follows:
The initial tank contains 390 tons of ballast at a VCG of 9.85 feet. The vertical moment for this ballast is 390 tons x 9.85 feet = 3,841.5 ft-long tons. After discharging 250 tons of ballast with a VCG of 12.0 feet, the remaining ballast is 390 tons - 250 tons = 140 tons, with a vertical moment of 140 tons x 9.85 feet = 1,379 ft-long tons. The difference between the initial vertical moment and the remaining vertical moment is 3,841.5 ft-long tons - 1,379 ft-long tons = 842 ft-long tons, which matches the correct answer A.
The other options are incorrect because they do not accurately represent the change in vertical moment after the ballast discharge.
Related Questions
#4336 On the DEEP DRILLER, among the starboard-side valves to open when deballasting from ballast tank 2S is valve _______________.#4337 The distance from the still water level (corrected for tides and storm surge) to the bottom of the hull of a MODU is the _______________.#4339 While loaded as shown in the COASTAL DRILLER Sample Load Form #3 (Drilling), 236 kips are discharged from 80 feet AF0 and 19.79 feet to port of the centerline. What is the resulting starboard leg reaction? #4340 On the DEEP DRILLER, when using the starboard #2 ballast pump to provide water to the main deck from the sea, it is necessary to open valves _______________.#4341 For a jack-up, the angle of maximum stability corresponds approximately to the angle of _______________.