The skater's final angular speed is equal to 12 rad/s.
When implemented to angular momentum, the regulation of conservation means that the momentum of a rotating item is no longer exchanged until some form of external torque is carried out. Torque, in this sense, can check with any outside pressure that acts upon the object for the purpose to twist or rotate.
The law of conservation of angular momentum states that once no external torque acts on an item, no trade of angular momentum will occur. The angular momentum of a machine is conserved as long as there may be no net external torque performing on the machine.
In angular kinematics, the conservation of angular momentum refers back to the tendency of a device to keep its rotational momentum inside the absence of outside torque. For a round orbit, the system for angular momentum is (mass) ×(pace) ×(radius of the circle): (angular momentum) = m × v × r.
Learn more about angular momentum here brainly.com/question/7538238
#SPJ4
Answer:
6400 m
Explanation:
You need to use the bulk modulus, K:
K = ρ dP/dρ
where ρ is density and P is pressure
Since ρ is changing by very little, we can say:
K ≈ ρ ΔP/Δρ
Therefore, solving for ΔP:
ΔP = K Δρ / ρ
We can calculate K from Young's modulus (E) and Poisson's ratio (ν):
K = E / (3 (1 - 2ν))
Substituting:
ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)
Before compression:
ρ = m / V
After compression:
ρ+Δρ = m / (V - 0.001 V)
ρ+Δρ = m / (0.999 V)
ρ+Δρ = ρ / 0.999
1 + (Δρ/ρ) = 1 / 0.999
Δρ/ρ = (1 / 0.999) - 1
Δρ/ρ = 0.001 / 0.999
Given:
E = 69 GPa = 69×10⁹ Pa
ν = 0.32
ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)
ΔP = 64.0×10⁶ Pa
If we assume seawater density is constant at 1027 kg/m³, then:
ρgh = P
(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa
h = 6350 m
Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.
Answer:
6.39 J of energy is needed to generate 0.71 * 10⁻¹⁶ kg mass
Explanation:
According to the Equation: E = mc²
where the mass, m = 0.71 * 10⁻¹⁶ kg
the speed of light, c = 3 * 10⁸ m/s
The amount of energy needed to generate a mass of 0.71 * 10⁻¹⁶ kg is calculated as follows:
E = (0.71 * 10⁻¹⁶) (3 * 10⁸)²
E = 0.71 * 10⁻¹⁶ * 9 * 10¹⁶
E = 0.71 * 9
E = 6.39 J
You may confuse two minerals by their colors so using streak is another way for showing their "true colors "