437x9
is ur answer. I'm not sure tho hope it helps
We're u can never put it back together
Answer:
the buoyant force on the chamber is F = 7000460 N
Explanation:
the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.
Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume
mass of water displaced = density of seawater * volume displaced
m= d * V , V = 4/3π* Rext³
the buoyant force is the weight of this volume of seawater
F = m * g = d * 4/3π* Rext³ * g
replacing values
F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N
Note:
when occupied the tension force on the cable is
T = F buoyant - F weight of chamber = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N
Answer:
w = 4,786 rad / s
, f = 0.76176 Hz
Explanation:
For this problem let's use the concept of angular momentum
L = I w
The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved
Initial Before sticking
L₀ = 0 + I₂ w₂
Final after coupling
= (I₁ + I₂) w
The moments of inertia of a disk with an axis of rotation in its center are
I = ½ M R²
How the moment is preserved
L₀ = 
I₂ w₂ = (I₁ + I₂) w
w = w₂ I₂ / (I₁ + I₂)
Let's reduce the units to the SI System
d₁ = 60 cm = 0.60 m
d₂ = 40 cm = 0.40 m
f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz
Angular velocity and frequency are related.
w₂ = 2 π f₂
w₂ = 2π 3.33
w₂ = 20.94 rad / s
Let's replace
w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)
w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)
Let's calculate
w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)
w = 20.94 1.28 / 5.6
w = 4,786 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 4.786 / 2π
f = 0.76176 Hz
The answer is convex image