1) What is the density of an object whose dry weight is 13N and whose weight sub

Question

1) What is the density of an object whose dry weight is 13N and whose weight submerged in water is 9.5N?

2) A sealed chamber, totally submerged, is anchored to the ocean floor. If the chamber has a mass of 2.5x 10^7kg and a volume of 2.8 x10^4 m^3;

a)What is the buoyant force on the chamber?

b)What is the tension in the anchor chain holding the chamber under water?

c) If the anchor chain were cut, what would be the acceleration of the chamber as it rises to the surface?

3) If an object with a specific gravity of 2.68 has a volume of 1.3m^3, what is the weight of th eobject and the buoyant force acting on the object if it is completely submerged.

Explanation / Answer

1.

Weight f the object = mg = oVog

13 N = oVog

Vo= 13/og -----------------(1)

Buoyant force = True weight – apparent weight

Fb = 13 N – 9.5 N = 3.5 N

wVwg = 3.5 -----------(2)

But by Archimedes principle

Vw = Vo = 13/og ------------(3)

Put (3) in (1)

w *(13/og )*g = 3.5

o =w*(13/3.5) = 1000*(13/3.5) = 3714.3 kg/m^3

2.

a) Fb = wVwg = wVog = 1027*2.8 x10^4*9.8 = 2.8*10^8 N

b) Refer below figure,

Applying Newton’s second law

Fb = Fg + T

T = Fb – Fg

T = Fb – m*g

T = 2.8*10^8 - 2.5*10^7*9.8 = 3.5*10^7 N

c) T=0 =>

Fnet = ma

Fb – m*g = ma

a = (Fb – m*g) = (3.5*10^7)/ (2.5*10^7) = 1.4 m/s^2

3. Specific density = density of the object / density of water

o = density of the object = Specific density * density of water = 2.68*1000 = 2680 kg/m^3

Weight = oVog = 2680*1.3*9.8 = 34143.2 N

Fb = wVog = 1000*1.3*9.8 = 12740 N

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.