Answer question 3 asap plz The center than the edge. Does the period of oscillat

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Answer question 3 asap plz

The center than the edge. Does the period of oscillation increase or decrease relative to that found in (c). A uniform object of density rho0 = 2.5 g/cm3 and volume V0 = 10 cm3 is completely submerged in a fluid of density rho = 3.8 g/cm3 at a depth h = 4 cm below the surface. The acceleration due to gravity is g = 10 m/s2. Find the force (magnitude and direction) that must be exerted on the object to hold it at the depth h The force found in (a) is removed Find the acceleration of the object Neglecting the effects of friction, determine how fast the object will be moving when it breaks the surface of the fluid Treating the object as a point mass, determine the maximum height to which the object will rise. A tube is constructed such that one end has cross section A1 = 5 cm2 and it narrows down to a cross section A2 = 2 cm2. The region with cross section A2 has a tee tube of cross section A4 = 0.1 cm2 projecting down a distance h = 7 cm into a fluid of density rho = 3 g/cm3. The fluid is otherwise open to the atmospheric pressure In what follows, assume the air is incompressible and has a density rho0 = 1 2 times 10-3 g/cm3 and the acceleration due to gravity is g Air enters A1 with speed V1 = 200 cm/s Determine the air pressure in region A1. Determine the air pressure in region A2 Determine the height to which the fluid rho will rise in tube At. A sphere of mass M = 3 Kg and radius R = 20 cm is free to rotate about an axie through its center (I = 2 / 5 MR2). The sphere is placed near a wall and

Explanation / Answer

a) the forces are gravity, the buoyant force, and the applied force. the balance is: Fapp + Fg = Fb Fb = (pf)Vg, where pf is the density of the fluid Fg = (po)Vg, where p0 is the density of the object Fapp = (pf)Vg - (po)Vg = (pf - po)Vg = (3.8 - 2.5)(10)(10)(0.001) = 0.13 N (downward since its positive and we assumed downward in the balance) the 0.001 is the conversion from grams to kilograms. b) Fb - Fg = ma, where upward is now the positive direction. (pf)Vg - (po)Vg = (po)Va a = (pf - po)g/po = (3.8 - 2.5)(10)/2.5 = 5.2 m/s^2 c) the kinematic equation is: vf^2 = v0^2 + 2ah = 0 + 2(5.2)(0.04) = 0.416 m^2/s^2 vf = 0.6450 m/s d) the final velocity in (c) becomes initial velocity, final velocity is 0 (maximum height), and acceleration is just gravity (left water) vf^2 = v0^2 + 2ah 0 = 0.416 - 2(10)h h = 0.0208 m = 2.08 cm

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