Answer:
The volume of water displaced by the raft is 0.233 m³
Explanation:
The question relates to Archimedes' principle which states that the buoyant force experienced by an object immersed in a fluid is equal to the weight of (the force of gravity on) the displaced fluid
The given parameters are;
The combined mass of the person and the raft, m = 233 kg
The liquid on which the raft is located = Water
The density of water, = 1000 kg/m³
Weight = Mass, m × g
Where;
m = The mass of the object
g = The acceleration due to gravity = 9.8 m/s²
Given that the raft is on the surface of the water (floating), the buoyant force is equal to the combined weight of the person and the raft = 233 kg
The combined weight of the person and the raft, = 233 kg × 9.8 m/s² = 2,283.4 N
Therefore;
The buoyant force = 2,283.4 N = The weight of the water displaced
The mass of the water displaced, , = 2,283.4 N/(9.8 m/s²) = 233 kg
Density = Mass/Volume
The volume of water displaced by the raft = The mass of the water displaced/(The density of the water) = 233 kg/(1,000 kg/m³) = 0.233 m³.
Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density
Drag force FD is given as
Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D
T = 16653.32 N