A 950-kg cylindrical can buoy floats vertically in sea-water. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when an 80.0-kg man stands on top of it.

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Raymond Foley · Accepted Answer

Given data:  - The mass of the buoy is: M=950kg.  - The diameter of the buoy is: D=0.900m.  - The radius of the buoy is: r=D2=0.450m.  - The mass of the man is: m=80.0kg.  - The density of seawater is: ρ=1.03×103kgm3.  Required data  We are asked to determine the additional distance the buoy will sink when the man stands on top of it.  Solution  Since the buoy is still floating after the man stands on it and according to Archimedes principle from Equation (1), the additional buoyant force exerted on the buoy, equal to weight of the additional fluid displaced by the man, is also equal to the additional weight of the man:  B=Wman  Substitute for B from Equation (1) and for W from Equation (2):  ρgVdisp=mg  ρVdisp=m  where Vdisp is the additional volume displaced when the man stands on the buoy. It is equal to AD where A=πr2 is the area of the buoy and d is the distance the buoy sink:  ρ(πr2)d=m  Solve for d:  d=mπr2ρ  Substitute numerical values:  d=80.0kgπ(0.450m)2(1.03×103kgm3)  =0.122m

Melissa Moore · Accepted Answer

Answer:   Additional depth is 0.13 m   Explanation:   mass of cylinder, M=950kg   Let the initial height inside water is h.   diameter =0.9m   radius, r=0.45m   mass of man, m=80kg   density of water =1000kgm3   Let the additional distance is y.   For initial stage:   Buoyant force = weight   Volume immersed×density of water×g=Mg   3.14×0.45×0.45×1000×h=950   h=1.49m   Now:   3.14×0.45×0.45×1000×(h+y)=950+80   950+635.85y=950+80   y=0.13m

karton · Accepted Answer

Step 1 Weight of an object, W=mg equation 1 where m=80kg,  g=9.8 m/s2 then Wm=(80kg)(9.8m/s2) =784N Cross sectional area of the cylindrical can is a=πd24 where, a is cross-sectional area of the buoy d is diameter then, a=π×(0.9m)24 =0.636m2 Step 2 Volume of the cylindrical can buoy which sin, V=ah then, V=(0.636m2)h We know that, Weight of the water displaced will be equal to the buoyant force then, BSW=ρwaterVg equation 2 here BSW is Buoyant force ρwater is the density of sea water. put all known values in equation 2 Bw=(1030kg/m3)((0.636m2)×h)(9.8ms−2) =6419.78×h N/m As of, water is replaced when man stands on top buoy, Wm=Bsw equation 3 put all known values in equation 3 ⇒784N=6419.78×h N/m h=0.122m Answer: Additional distance through which the buoy sink is 0.122m.

A 950-kg cylindrical can buoy floats vertically in sea-water. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when an 80.0-kg man stands on top of it.

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