A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft must hold a 730 kg steel cube with edges of length 45.0 cm without sinking. Assume the density of water in the pool is 1000 kg/m3.a. The students use a crane to gently place the cube on the raft but accidentally place it off center. The cube remains on the raft for a few moments and then the raft tilts, causing the cube to slide off and sink to the bottom of the pool. Theraft remains floating in the pool. In a coherent paragraph-length response, indicate whether the water level in the pool when the cube is on the bottom of the pool is higher than, lower than, or the same as when the cube is on the raft, and explain your reasoning. For both cases, assume that there is no motion of the water. b.The bottom of the pool is 5 m below the surface of the water. The crane is now used to lift the cube out of the pool. The cube is lifted upward at a slow and constant speed, soall drag forces are negligible. i.Predict how the force exerted by the crane on the cube when the bottom of the cube is at a depth of 2.0 m compares to the force exerted by the crane on the cube when the bottom of the cube is at a depth of 4.0 m. Explain your reasoning.

(a) The water level in the pool when the cube is on the bottom of the pool is the same as when the cube is on the raft

(b) The force exerted by the crane on the cube is the same in both cases

Explanation:

a. Here we note that Archimedes's principle states that for a body immersed, either wholly or partial in a fluid, the buoyancy or upward force acting on the body in the fluid is equal to the wight of the displaced fluid. Since the raft and the cube form a system experiencing an upward force equal to their weight, then, for the raft to float, the volume displaced is equal to the weight of the raft plus the metal cube. When the cube sinks the fluid, there is less weight on the raft, so it rises while the water level remain  the same as the proportionate amount of volume is taken up by the cube and the floating raft.

b. Where the bottom of the cube is at a depth of 2.0 m we have a buoyant force acting from the bottom of the cube and weight of  the fluid over head acting from the top

Buoyant force - Weight of cube - weight of 2.0 m column of water above + Upward force due to pressure acting from beneath the cube at (2+0.45 m)

When the cube is at 4.0 m we have the buoyant we have

Buoyant force - Weight of cube - weight of 4.0 m column of water above + Upward force due to pressure acting from beneath the cube at (4+0.45 m)

Therefore the force exerted by the crane on the cube is constant in both cases.

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day