Answer:
rotation
brooo like ittttttttttttt
Answer:
the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.
Explanation:
This is a fluid mechanics problem, where as the boat is in equilibrium with the pushing force we can write Newton's second law
B- W = 0
B = W
the thrust force is equal to the weight of the liquid that is dislodged
B = ρ g V
we substitute
ρ g V = m g
V = m /ρ_fluid 1
we can write the mass of the pot as a function of its density
ρ_body = m / V_body
m = ρ_body V_body
V_fluid / V_body = ρ_body / ρ _fluid 2
Equations 1 and 2 are similar, although 2 is easier to analyze, the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.
The effect appears the pot as if it had a lower apparent weight
Answer:
ω = 0.467 rad/s
Explanation:
given,
tangential force exerted by the person = 37.7 N
radius of merry-go-round = 2.75 m
mass of merry-go-round = 144 Kg
angle = 33.2°
moment of inertia
I = 544.5 kg.m²
torque = force x radius
τ = 37.7 x 2.75
τ = 103.675 N.m
angular acceleration


α = 0.190 rad/s²
now
,
distance = 
d = 0.579 rad
we know,
using equation of rotational motion
t = 2.46 s
angular speed
ω = α x t
ω = 0.19 x 2.46
ω = 0.467 rad/s
Given,
The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m
The initial speed of the water, v₁=12.5 m/s
The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m
From the continuity equation,

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.
On substituting the known values,

Thus, the flow rate of the water at the later position is 5.99 m/s
Abyssal plain
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