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horsena [70]
2 years ago
7

Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence

of dark energy. Rank each model from left to right based on the ratio of its actual mass density to the critical density, from smallest ratio (mass density much smaller than critical density) to largest ratio (mass density much greater than critical density).Smallest to largesta.coasting universeb.critical universec.recollapsing universe
Physics
1 answer:
sineoko [7]2 years ago
7 0

Answer:

From smallest ratio to the largest ratio:

Coasting Universe - Critical Universe - Recollapsing Universe(From left to right)

Explanation:

The coasting universe is one that expands at a constant rate given by the Hubble constant throughout all of cosmic time. It has a ratio of actual density to critical density that is less than 1

The critical universe is one that is at balance with no expansion .I.e. the actual density and the critical density are equal, which makes the ratio of actual density to critical density to be equal to 1

Recollapsing Universe: The expansion of the universe reverses in the future and the universe eventually recollapses. The recollapsing universe has the ratio of the actual density to the critical density to be greater than 1

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The key difference between the binomial and hypergeometric distribution is that
Ivenika [448]

Explanation:

Both distributions describe the number of times an event occurs in a givn number of trials. In the binomial distribution, the probability is the same for each trial. While in the hypergeometric distribution, each trial changes the probability of each subsequent trial, since there is no replacement.

5 0
2 years ago
A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.30 m/s . The coefIficient o
Artemon [7]

Answer:

Explanation:

Mass =11.2kg

Constant velocity =3.3m/s

μk=0.25

Since the body is moving in constant velocity, then the acceleration is zero(0).

ΣF = Σ(ma)

The normal force acting on the body is upward and the weight is acting downward

Then ΣFy=0

Therefore, N=W

W=mg=11.2×9.8=109.76N

So, N=W=109.76N

Frictional force is given as

Fr=μkN

Fr=0.25×109.76

Fr=27.44N

Frictional force acting against the motion is 27.44N

Then the forward force moving the body forward

ΣF = Σ(ma)

Since a = 0

Then,

ΣF = 0

F-Fr=0

Then F=Fr

So the force moving the body forward is 27.44N

8 0
3 years ago
What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
maks197457 [2]

Answer:

Wavelength = 1.36 * 10^{-34} meters

Explanation:

Given the following data;

Mass = 0.113 kg

Velocity = 43 m/s

To find the wavelength, we would use the De Broglie's wave equation.

Mathematically, it is given by the formula;

Wavelength = \frac {h}{mv}

Where;

h represents Planck’s constant.

m represents the mass of the particle.

v represents the velocity of the particle.

We know that Planck’s constant = 6.6262 * 10^{-34} Js

Substituting into the formula, we have;

Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43}

Wavelength = \frac {6.6262 * 10^{-34}}{4.859}

Wavelength = 1.36 * 10^{-34} meters

7 0
2 years ago
Guys help me with this question.​
sleet_krkn [62]

Answer:

Explanation:

Brownian motion is a random (irregular) motion of particles e.g smoke particle. The set up in the diagram can be used to observe the motion of smoke.

1. The apparatus used are:

A is a source of light

B is a converging lens

C is a glass smoke cell

D is a microscope

2. The uses of the apparatus are:

A - produces the light required to so as to see clearly the movement of the particles.

B - converges the rays of light from the source to the smoke cell.

C - is made of glass and used for encamping the smoke particles so as not to mix with air.

D - is used for the clear view or observation or study of the motion of the smoke particles in the cell.

4 0
3 years ago
Find the ratio of the new/old periods of a pendulum if the pendulum were transported from earth to the moon, where the accelerat
vichka [17]
The period of a pendulum is given by
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length and g is the gravitational acceleration.

We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
\frac{T_m}{T_e} =  \frac{2 \pi  \sqrt{ \frac{L}{g_m} } }{2 \pi  \sqrt{ \frac{L}{g_e} } }=    \sqrt{ \frac{g_e}{g_m} }
where the labels m and e refer to "Moon" and "Earth".

Since the gravitational acceleration on Earth is g_e = 9.81 m/s^2 while on the Moon is g_m=1.63 m/s^2, the ratio between the period on the Moon and on Earth is
\frac{T_m}{T_e}= \sqrt{ \frac{g_e}{g_m} }= \sqrt{ \frac{9.81 m/s^2}{1.63 m/s^2} }=2.45

3 0
2 years ago
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