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∙ 16y agoJust divide a boat's sail area in square feet by its wetted surface area in square feet (SA/WS = SA ÷ WS),
Abdu Khalifi
water displacement
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1. Calculate the surface area 2. Calculate the volume 3. Divide
You calculate the areas of two shapes and then divide one area by the other to find the ratio of their areas.
Multiply the surface area by the volume Also stop cheating ur hw
water displacement
1) Calculate the area 2) Calculate the volume 3) Divide the area by the volume to get the ratio
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SA/D Ratio is the sail area/displacement ratio. This ratio indicates how fast the boat is in light wind. The higher the number the faster the boat.* Cruising Boats have ratios between 10 and 15.* Cruiser-Racers have ratios between 16-20.* Racers have ratios above 20.* High-Performance Racers have ratios above 24.SA / D = Sail Area / (Displacement in Cubic Feet )2/3
reduction ratio= initial cross sectional area/final cross sectional area
You measure or calculate the surface area; you measure or calculate the volume and then you divide the first by the second. The surface areas and volumes will, obviously, depend on the shape.
How to calculate the ratio of the inlet-to-exit area of the nozzle
1. Calculate the surface area 2. Calculate the volume 3. Divide
To find the displacement from a negative velocity-time graph, you need to calculate the area under the curve for the portion representing displacement. If the velocity is negative, the displacement will be in the opposite direction. The magnitude of the displacement is equal to the absolute value of the area under the curve.
FAR=Total floor area of building / Total lot area
The surface-area-to-volume ratio may be calculated as follows: -- Find the surface area of the shape. -- Find the volume of the shape. -- Divide the surface area by the volume. The quotient is the surface-area-to-volume ratio.
To calculate the surface area to volume ratio, simply divide the surface area of the object by its volume. This ratio is commonly used in science to understand how efficiently an object exchanges materials with its environment, with a higher ratio indicating better surface area for exchange relative to its volume.