Answer:
v = 15 m / s
Explanation:
In this exercise we are given the position function
x = 5 t²
and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval
let's look for the displacements
t = 0 x₀ = 0 m
t = 3 = 5 3 2
x_{f} = 45 m
we substitute
v = 15 m / s
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.