Answer:
1. Least massive stars are the coolest and least luminous, lower right of main sequence, on HR diagram.
2. Most massive are the hottest and most luminous, upper left of main sequence on Hr Diagram.
3. The radius of stars are related to their sprectral type. having the O being the hottest upper left and M being the coolest bottom right.
To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.
PART A) Normal Force.


Here,
Normal reaction of the ring is N and velocity of the ring is v




PART B) Acceleration





Negative symbol indicates deceleration.
<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>
Answer:
a = 3.27 m/s²
v = 2.56 m/s
Explanation:
given,
mass A = 1 kg
mass B = 2 kg
vertical distance between them = 1 m










a = 3.27 m/s²
The speed of the system at that moment is:
v² = u² + 2×a×s
v² = 0² + 2× 3.27 × 1
v ² = 6.54
v = 2.56 m/s
Answer:
R = 98304.75 m = 98.3 km
Explanation:
The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.
Density = Mass/Volume
Now, it is given that the density of Earth has become:
Density = 1 x 10⁹ kg/m³
Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg
Volume = 4/3πR³ (Volume of Sphere)
R = Radius of Earth = ?
Therefore,
1 x 10⁹ kg/m³ = (5.97 x 10²⁴ kg)/[4/3πR³]
4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)
R³ = (3/4)(5.97 x 10¹⁵ m³)/π
R = ∛[0.95 x 10¹⁵ m³]
<u>R = 98304.75 m = 98.3 km</u>