Answer:
ω = ω₀ + α t
ω² = ω₀² + 2 α θ
θ = θ₀ + ω₀ t + ½ α t²
Explanation:
Rotational kinematics can be treated as equivalent to linear kinematics, for this change the displacement will change to the angular displacement, the velocity to the angular velocity and the acceleration to the angular relation, that is
x → θ
v → ω
a → α
with these changes the three linear kinematics relations change to
ω = ω₀ + α t
ω² = ω₀² + 2 α θ
θ = θ₀ + ω₀ t + ½ α t²
where it should be clarified that to use these equations the angles must be measured in radians
Where speed is distance/time, velocity is displacement/time.
What this means is that velocity is the length covered in relation to the starting point.
Speed is just the distance travelled no matter where you began.
When going around a circular track, you might have a speed value. However, since you get back to the same location at every lap, you have 0 velocity.
Hope I helped :)
The particle in the air match the mass with the particles around the balloon stoping it from floating any higher
I think this one's B. energy and work are both measured in joules.
Answer:
F = 1.24*10^4 N
Explanation:
Given
Depth of the ship, h = 25 m
Density of water, ρ = 1.03*10^3 kg/m³
Diameter of the hatch, d = 0.25 m
Pressure of air, P(air) = 1 atm
Pressure of water =
P(w) = ρgh
P(w) = 1.03*10^3 * 9.8 * 25
P(w) = 2.52*10^5 N/m²
P(net) = P(w) + P(air) - P(air)
P(net) = P(w)
P(net) = 2.52*10^5 N/m²
Remember,
Pressure = Force / Area, so
Force = Area * Pressure
Area = πr² = πd²/4
Area = 3.142 * 0.25²/4
Area = 3.142 * 0.015625
Area = 0.0491 m²
Force = 0.0491 * 2.52*10^5
F = 12373 N
F = 1.24*10^4 N
