Well one of the factors is that the species is poorly adapted to the their environment and die off
Answer:
The required angular speed the neutron star is 10992.32 rad/s
Explanation:
Given the data in the question;
mass of the sun M = 1.99 × 10³⁰ kg
Mass of the neutron star
M = 2( M )
M = 2( 1.99 × 10³⁰ kg )
M = ( 3.98 × 10³⁰ kg )
Radius of neutron star R = 13.0 km = 13 × 10³ m
Now, let mass of a small object on the neutron star be m
angular speed be ω.
During rotational motion, the gravitational force on the object supplies the necessary centripetal force.
GmM = / R² = mRω²
ω² = GM = / R³
ω = √(GM = / R³)
we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²
we substitute
ω = √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)
ω = √( 2.65466 × 10²⁰ / 2.197 × 10¹²
ω = √ 120831133.3636777
ω = 10992.32 rad/s
Therefore, The required angular speed the neutron star is 10992.32 rad/s
Explanation:
Sorry but I don't Understand question
The definition of a scale of 1: 166 will mean that the scale of 1 in the model will be equivalent to 166 times the measurement in the real model, therefore we will have that the height would be 166 times smaller than the 179m given:
The same for the diameter,
The volume of a cylinder is given as
Therefore the volume would be
Answer:
307.07N
Explanation:
Given
Weight of object = 250N
angle of elevation = 18.0°
coefficient of kinetic friction n = 0.26
acceleration = 9.81m/s^2
Required
Force parallel to the ramp
Using Newton second law
F = ma
Fm - Ff = ma
Fm - nR = ma
Fm = ma +nR
Fm = 25(9.81) + 0.26(Wcostheta)
Fm = 25(9.81) + 0.26(250cos18)
Fm = 245.25+61.81
Fm = 307.07N
Hence he magnitude of the force that the rope exerts on the crate parallel to the ramp is 307.07N