Answer:
L= 0.059 mH
Explanation:
Given that
R = 855 Ω and C = 6.25 μF
V= 84 V
Frequency
ω = 51900 1/s
We know that
L=Inductance
C=Capacitance
ω =angular Frequency
ω² L C =1
(51900)² x L x 6.25 x 10⁻⁶ = 1
L= 5.99 x 10⁻⁵ H
L= 0.059 mH
Max (15 kg) and maya (12 kg) are riding on a merry-go-round that rotates at a constant speed. max is sitting on the edge of the
merry-go-round, 2.4 m from the center, and maya is sitting somewhere between the edge and the center. the merry-go-round is a solid disk with a radius of 2.4 m, a mass of 230 kg, and is rotating at a constant rate of 0.75 rev/s. the system, which includes max, maya, and the merry-go-round, has 8700 j of rotational kinetic energy. how far away is maya from the center of the merry-go-round?
1 answer:
Refer to the diagram shown below.
M = 230 kg, the mass of the wheel
m₁ = 15 kg, Max's mass
m₂ = 12 kg, Maya's mass
r = 2.4 m, the radius of the wheel
Max is located at 2.4 m from the center of the weel.
Let x = the distance of Maya from the center of the wheel.
The polar moment of inertias of the system is
J = (1/2)Mr² (for the wheel)
+ m₁r² (for Max)
+ m₂x² (for Maya)
That is,
J = 0.5*(230)*(2.4)² + (15)*(2.4)² + (12)x² kg-m²
= 748.8 + 12x² kg-m²
The angular velocity is
ω = 0.75 rev/s = (0.75*2π) rad/s = 4.7124 rad/s
By definition, the rotational KE is
KE = (1/2)*J*ω²
Because the rotational KE is 8700 J, therefore
(1/2)*(748.8 + 12x²)*(4.7124)² = 8700
11.1034(748.8 + 12x²) = 8700
748.8 + 12x² = 783.547
12x² = 34.747
x = 1.702 m
Answer: 1.7 m (nearest tenth)
M = 230 kg, the mass of the wheel
m₁ = 15 kg, Max's mass
m₂ = 12 kg, Maya's mass
r = 2.4 m, the radius of the wheel
Max is located at 2.4 m from the center of the weel.
Let x = the distance of Maya from the center of the wheel.
The polar moment of inertias of the system is
J = (1/2)Mr² (for the wheel)
+ m₁r² (for Max)
+ m₂x² (for Maya)
That is,
J = 0.5*(230)*(2.4)² + (15)*(2.4)² + (12)x² kg-m²
= 748.8 + 12x² kg-m²
The angular velocity is
ω = 0.75 rev/s = (0.75*2π) rad/s = 4.7124 rad/s
By definition, the rotational KE is
KE = (1/2)*J*ω²
Because the rotational KE is 8700 J, therefore
(1/2)*(748.8 + 12x²)*(4.7124)² = 8700
11.1034(748.8 + 12x²) = 8700
748.8 + 12x² = 783.547
12x² = 34.747
x = 1.702 m
Answer: 1.7 m (nearest tenth)
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