Asked by Justin

A flat-bottom river barge is 30.0 ft wide, 85.0 ft long, and 15.0 ft deep. (a) How many ft3 of water will it displace while the top stays 3.00 ft above the water? (b) What load in tons will the barge contain under these conditions if the empty barge weighs 160 tons in dry dock?

Answers

Answered by Henry
a. Vw = L*W*h = 85 *30 * (15-3) = 30,600 Ft^3.

b. 30,600Ft^3 * 62.4Lbs/Ft^3 = 1.909,440 Lbs. = 955 Tons = Wt. of water displaced.

Load = 955 - 160 = 795 Tons.

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