The distances to the most of these galaxies estimated from their red shifts and the application of Hubble's Law.
<h3>What is Hubble's Law?</h3>
The discovery in physical cosmology that galaxies are travelling away from Earth at rates proportionate to their distance is known as Hubble's law, often referred to as the Hubble-Lemaître law or Lemaître's law. In other words, they are travelling away from Earth more quickly the more away they are. The redshift of the galaxies—a shift in the light they produce toward the red end of the visible spectrum—has been used to calculate their velocities.
The Sloan Digital Sky Survey includes many thousands of galaxies in its spectroscopic catalogs. The distances to the most of these galaxies estimated
from their red shifts and the application of Hubble's Lawh
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Answer:
The net charge on the shell is 30x10^-9C
Explanation:
Pls see attached file
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
The forces acting on the book (Fn and Fg) are equal. Therefore, the netforce is zero and it is an unbalanced force.
Answer:
D) Q/2
Explanation:
The relationship between charge Q, capacitance C and voltage drop V across a capacitor is
(1)
In the first part of the problem, we have that the charge stored on the capacitor is Q, when the voltage supplied is V. The capacitance of the parallel-plate capacitor is given by
where 
is the vacuum permittivity, A is the area of the plates, d is the separation between the plates.
Later, the voltage of the battery is kept constant, V, while the separation between the plates of the capacitor is doubled: 
. The capacitance becomes
And therefore, the new charge stored on the capacitor will be
