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3 years ago

6

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

of dark energy:
A) recollapsing universe
B) coasting universe
C) critical universe
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).

1 answer:

Keith_Richards [23]3 years ago

3 0

Answer:

1. Recollapsing universe

2. Critical universe

3. Coasting universe

Explanation:

Recollapsing universe has dark matter density greater than critical density. While critical universe has its matter density equal to the critical sensity. Coasting universe on the other hand has much smaller matter density compared to critical density.

Note that the critical density is approximately 10^-20 grams/cm3

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2. A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotatin

ycow [4]

Answer:

$\frac{0.065}{r}$

Explanation:

The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);

$v=\sqrt{\mu gr}....................(1)$

where $\mu$ is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.

Given;

v = 0.8m/s

g = $9.81m/s^2$

r = ?

$\mu=?$

In order to solve for $\mu$ , we can simply make it the subject of formula from equation (1) as follows;

$v^2=\mu gr\\hence\\\mu=\frac{v^2}{gr}.................(2)$

since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.

Therefore;

$\mu=\frac{0.8^2}{9.81r}$

$\mu=\frac{0.64}{9.81r}\\\\\mu=\frac{0.065}{r}$

8 0

3 years ago

A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon

Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

$n=\frac{m}{M_m}$

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

$n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol$

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

$N=n N_A$

where

n is the number of moles

$N_A=6.022\cdot 10^{23} mol^{-1}$ is the Avogadro number

Substituting,

$N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18}$ atoms

The energy emitted by each atom (the energy of one photon) is

$E_1 = \frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda=508 nm=5.08\cdot 10^{-7}nm$ is the wavelength

Substituting,

$E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J$

And so, the total energy emitted by the sample is

$E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J$

4 0

3 years ago

41. Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 h

romanna [79]

Answer:

138.3 days

Explanation:

Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.

The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.

T^2 is proportional to r^3

That is,

T^2/r^3 = constant

98^2 / 62^3 = T^2 / 78^3

Make T^2 the subject of formula.

T^2 = 98^2 / 62^3 × 78^3

T^2 = 19123.2

T = sqrt ( 19123.2 )

T = 138.2867 days

Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.

4 0

3 years ago

John is traveling north at 20 meters/second and his friend Betty is traveling south at 20 meters/second. If north is the positiv

Tpy6a [65]

I think it's just 20/20

3 0

3 years ago

Read 2 more answers

Red laser light from a He-Ne laser (λ = 632.8 nm) creates a second-order fringe at 53.2° after passing through the grating. What

Svetlanka [38]

Explanation:

It is given that,

Wavelength of red laser light, $\lambda=632.8\ nm=632.8\times 10^{-9}\ m$

The second order fringe is formed at an angle of, $\theta=53.2^{\circ}$

For diffraction grating,

$d\ sin\theta=n\lambda$

$d=\dfrac{n\lambda}{sin\theta}$ , n = 2

$d=\dfrac{2\times 632.8\times 10^{-9}}{sin(53.2)}$

$d=1.58\times 10^{-6}\ m$

The wavelength λ of light that creates a first-order fringe at 22 is given by :

$\lambda=d\ sin\theta$

$\lambda=1.58\times 10^{-6}\ sin(22)$

$\lambda=5.91\times 10^{-7}\ m$

$\lambda=591\ nm$

Hence, this is the required solution.

6 0

4 years ago