Answer:
a basis or standard for evaluation,assessment or comparison
Explanation:
a point use to find or describe the location of something.
Answer:
Actually, Polaris, also named alpha Ursa Minoris, is the brightest star in the Little Dipper. It marks the end of the handle. By a twist of luck, it also happens to reside very close to the North Celestial Pole (NCP). This is the point in the sky that all the stars in the north rotate around. It’s not exactly on the NCP, in fact it’s more than a Moons width away, so it scribes out a very small circle in long exposure star trail images like this one below. To the unaided eye it appears that all the stars rotate around Polaris while it remains fixed in one spot. During the last half of the 20th century Polaris’ variations had dropped to approximately 2%. No other Cepheid is known to have gone through this. Astronomers believed they were witnessing the evolution of the star before their very eyes, and that eventually we would see Polaris’ variations snuff out entirely.
Explanation:
Answer:
1) a) I₁ = 0.2941 kg m², b) I₂ = 0.2963 kg m², c) I_{total} = 0.5904 kg m²
3) α = 6.31 10⁶ rad / s²
Explanation:
1) The moment of inertia for bodies with high symmetry is tabulated, for a divo with an axis passing through its center is
I = ½ m r²
a) moment of inertia of the upper disk
I₁ = ½ m₁ r₁²
I₁ = ½ 1,468 0.633²
I₁ = 0.2941 kg m²
b) upper aluminum disc moment of inertia
I₂ = ½ m₂ r₂²
I₂ = ½ 1.479 0.633²
I₂ = 0.2963 kg m²
c) the moment of inertia is an additive scalar quantity therefore
I_{total} = I₁ + I₂
I_{total} = 0.2941 + 0.2963
I_{total} = 0.5904 kg m²
3) ask the value of the angular acceleration, that is, the second derivative of the angle with respect to time squared
indicate the angular velocity of the system w = 400 rev / s
Let's reduce the SI system
w = 400 rev / s (2π rad / rev) = 2513.27 rad / s
as the system is rotating we can calculate the centripetal acceleration
a = w² R
a = 2513.27² 0.633
a = 3.998 10⁶ m / s²
the linear and angular variable are related
a = α r
α = a / r
α = 3.998 10⁶ / 0.633
α = 6.31 10⁶ rad / s²
Answer:
the magnitude of the momentum of the two-ball system immediately after collision is 32.31 kg.m/s
Explanation:
Given;
mass of the first ball, m₁ = 1.0 kg
mass of the second ball, m₂ = 2.0 kg
initial velocity of the first ball, v₁ = 30 m/s due west
initial velocity of the second ball, v₂ = 6 m/s due north
From the principle of conservation of linear momentum;
the total momentum before collision = total momentum after collision
The westward momentum of the first ball, = m₁v₁ = 1 x 30 = 30 kg.m/s
The northward momentum of the second ball = m₂v₂ = 2 x 6 = 12 kg.m/s
The resultant momentum of the two balls;
R² = 30² + 12²
R² = 1044
R = √1044
R = 32.31 kg.m/s
Therefore, the magnitude of the momentum of the two-ball system immediately after collision is 32.31 kg.m/s