E. Galaxy Cluster
Explanation:
A galaxy cluster, or cluster of galaxies, is a structure that consists of anywhere from hundreds to thousands of galaxies that are bound together by mutual gravity.
A megaparsec is a million parsecs and there are about 3.3 light years in a mega-parsec. Parsec is rather a natural distance unit for astronomers. The standard abbreviation of a mega-parsec is Mpc.
A parsec is approximately 3.09 x 1016 meters, a megaparsec is about 3.09 x 1022 meters.
Hence, 8 megaparsecs is gigantic size and that can be only of a galaxy cluster consisting of hundreds and thousands of galaxies bounded together.
Keywords: galaxies, parsec, megaparsec, galaxy cluster
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<h2>
Answer:</h2>
(a) 10N
<h2>
Explanation:</h2>
The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force
is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F - 
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F -
= m x a
F -
= 50 x 0
F -
= 0
F =
-------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force
is opposing the movement of the box.
∑F = 1.5F - 
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F -
= m x a
1.5F -
= 50 x 0.1
1.5F -
= 5 ---------------------(iii)
<em>Substitute </em>
<em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
<em />
Answer:
331.75 V
Explanation:
Given:
Number of turns of the coil, N = 40 turns
Area, A = 0.06 m²
Magnetic Field, B = 0.4 T
Frequency, f = 55 Hz
Maximum induce emf, E₀ = NABω
but ω = 2πf
Maximum induce emf, E₀ = NAB(2πf₀)
Maximum induce emf, E₀ = 2πNABf₀
Where;
N is number of turns of the coil
A is area
B is magnetic field
ω is the angular velocity
f is the frequency
E₀ = 2 × π × 40 × 0.06 × 0.4 × 55
E₀ = 342.81 V
The maximum induced emf is 331.75 V
Answer:
56
Explanation:
I just want the points to be completely honest with you.
Answer:
v₂ = 306.12 m/s
Explanation:
We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:
A₁v₁ = A₂v₂
where,
A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²
v₁ = entrance velocity = 3 m/s
A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²
v₂ = exit velocity = ?
Therefore,
(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂
v₂ = (0.006 m³/s)/(0.0000196 m²)
<u>v₂ = 306.12 m/s</u>