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.
They help scientists explain concepts that are difficult to observe.
Answer:
c. vf is greator than v2, but less than v1
Explanation:
The principle of conservation of linear momentum states that when two or more bodies act upon one another, their total momentum remains constant.
In a system of colliding bodies the total momentum of the system just before the collision is the same as the total momentum just after the collision.
Collisions in which the kinetic energy is conserved are called elastic collision.
Collisions in which the kinetic energy is not conserved are called inelastic collisions. If the two objects stick together after the collision and move with a common velocity, the collision is said to be perfectly inelastic.
<em>The above scenario is a perfectly inelastic collision. The initial velocity of particle 1 was greater than particle 2 before collision. After collision, its velocity will reduce to a final velocity vf as it transfers some of its kinetic energy to particle 2; whereas, the velocity of particle 2 will increase to a final velocity vf as it absorbs some of the kinetic energy of particle 1.</em>
Therefore,
a. vf = v2 is wrong because vf is greater than v2
b. vf is less than v2 is wrong because vf is greater than v2
c. vf is greater than v2, but less than v1 is correct.
d. vf = v1 is wrong because vf is less than v1
Answer:
a. 0.342 kg-m² b. 2.0728 kg-m²
Explanation:
a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m
I = 1/2MR²
= 1/2 × 56.5 kg × (0.11 m)²
= 0.342 kgm²
So the moment of inertia of the skater is
b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)
I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m
I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²
= 0.1802 kg-m² + 0.6852 kg-m²
= 0.8654 kg-m²
The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².
So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²
