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Anettt [7]
3 years ago
14

If you were to divide the presentday universe into cubes whose sides are 10 million lightyears long, each cube would contain, on

average, about one galaxy similar in size to the Milky Way. Now suppose you travel back in time, to an era when the average distance between galaxies is 0.26 of its current value, corresponding to a cosmological redshift z = 2.7. How many galaxies similar in size to the Milky Way would you expect to find, on average, in cubes of that same size?
Physics
1 answer:
vovikov84 [41]3 years ago
8 0

Answer:

Explanation:

density of galaxies would be \frac{1}{0.27^3} times higher which is equal to 50.81.

It means  in a cube that today contains one galaxy the size of the Milky Way, we would instead find 50.81 galaxies this size.

You can round this off to 52

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Please help<br>Will give the brainliest!​
Ronch [10]

Answer:

both answer is option C

Explanation:

tag me brainliest

7 0
2 years ago
A certain fuel-efficient hybrid car gets gasoline mileage of 55.0 mpg (miles per gallon). (a) If you are driving this car in Eur
ololo11 [35]

We will start by defining the units and their respective equivalences between the proposed measurement systems

1km = 0.6214mi

1gallon = 3.788 litres

PART A ) The mileage of the car is 55mpg (Miles per gallon)

55mpg = 55(\frac{miles}{gallon}) (\frac{1km}{0.6214miles})(\frac{1gallon}{3.788L})

55mpg = 23.4km/L

Therefore the mileage of the car is 23.4km/L

PART B ) The mileage of the car means that the car travels 23.4km and consumes 1 liter of fuel. Then

1L = 23.4km

For the car to travel 1500km the amount of fuel would be,

1500km= (1500km)(\frac{1L}{23.4km})

1500km = 64.1L

But 1 gas tank can only hold 45Liters of fuel, then the number of tank required would be

\text{Number of tanks required} = \frac{64.1L}{45L}

\text{Number of tanks required} = 1.4tanks

Thus the number of tanks of gas required to drive 1500km is 1.4

6 0
3 years ago
A large, 34.0 kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of m
Lorico [155]

Answer:

length L of the clapper rod for the bell to ring silently = 0.756m

Explanation:

We are given;

Mass of Bell;m_b = 34 kg

Distance of centre of mass from pivot;d = 0.7m

The bells moment of inertia about an axis at the pivot;I = 18 kg.m²

Mass of clapper;m_c = 1.8 kg

Length of slender rod is L

Now, the formula for period of physical pendulum having small amplitude is given as;

T_b = 2π√(I/mgd)

Where;

I is moment of inertia

m is mass

g is acceleration due to gravity = 9.8 m/s²

d is distance from rotation axis to centre of gravity

Plugging in the relevant values and using mass of bell, we have;

T_b = 2π√(18/(34*9.81*0.7)

T_b = 2π√(18/(34*9.81*0.7)

T_b = 1.745 s

Now, the formula for period for a simple pendulum which is essentially what the clapper rod is would be;

T_c = 2π√(L/g)

Now, we want to find length of clapper L.

Thus, let's make it the subject;

L = g(T_c/2π)²

Now, we are told that for the bell to ring silently, T_b = T_c.

Thus, T_c = 1.745 s.

So,

L = 9.8(1.745/2π)²

L = 0.756m

7 0
3 years ago
How is the control group and experimental group different
zzz [600]
The control group is the independent variable and the experimental group is the dependent due to change during the experiment. The experimental group will usually rely on another variable in the experiment for change.
4 0
3 years ago
A horizontal spring is attached to a wall at one end and a mass at the other. The mass rests on a frictionless surface. You pull
Kaylis [27]

Answer:

 Δt'/ T% = 90.3%

Explanation:

Simple harmonic movement is described by the expression

         x = A cos (wt)

we find the time for the two points of motion

x = - 0.3 A

        -0.3 A = A cos (w t₁)

         w t₁ = cos -1 (-0.3)

         

remember that angles are in radians

        w t₁ = 1.875 rad

x = 0.3 A

        0.3 A = A cos w t₂

        w t₂ = cos -1 (0.3)

         w t₂ = 1,266 rad

         

Now let's calculate the time of a complete period

x= -A

        w t₃ = cos⁻¹ (-1)

        w t₃ = π rad

this angle for the forward movement and the same time for the return movement in the oscillation to the same point, which is the definition of period

         T = 2 t₃

         T = 2π / w     s

now we can calculate the fraction of time in the given time interval

        Δt / T = (t₁ -t₂) / T

        Δt / T = (1,875 - 1,266) / 2pi

        Δt / T = 0.0969

 

This is the fraction for when the mass is from 0 to 0.3, for regions of oscillation of greater amplitude the fraction is

         Δt'/ T = 1 - 0.0969

         Δt '/ T = 0.903

         Δt'/ T% = 90.3%

4 0
3 years ago
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