Answer:
298rad/s and 116.96 ohms
Explanation:
Given an L-R-C series circuit where
L = 0.450 H,
C=2.50×10^−5F, and resistance R= 0
In this situation we have a simple LC circuit with angular frequency
Wo = 1√LC
= 1/√(0.450)(2.50×10^-5)
= 1/√0.00001125
= 1/0.003354
= 298rad/s
B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.
Wi/W° = (100-10)/100
Wi/W° = 90/100
Wi/W° = 0.90 ............... 1
Angular frequency of oscillation
The complete aspect of the solution is attached, please check.
Answer:
7.22 × 10²⁹ kg
Explanation:
For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.
So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km
The centripetal force F' = mRω² where R = radius of neutron star and ω = angular speed of neutron star
So, since F = F'
GMm/R² = mRω²
GM = R³ω²
M = R³ω²/G
Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²
Substituting the values of the variables into M, we have
M = R³ω²/G
M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²
M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²
M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²
M = 7217.66 × 10²⁶ kg
M = 7.21766 × 10²⁹ kg
M ≅ 7.22 × 10²⁹ kg
This correct answer would be C
