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2 years ago

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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).
(A) coasting universe
(B) critical universe
(C) recollapsing universe

1 answer:

Lynna [10]2 years ago

6 0

Answer:

1) Recollapsing universe

2) critical universe

3) Coasting universe

Explanation:

According to the smallest ration (ratio actual mass density to current density) to largest ration, rank of models for expansion of universe are

1) Recollapsing universe -in this, metric expansion of space is reverse and universe recollapses.

2) critical universe - in this, expansion of universe is very low.

3) Coasting universe - in this, expansion of universe is steady and uniform

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An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,

Sedbober [7]

Answer:

$a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa$

Explanation:

a. Internal energy and the relative specific volume at $s_1$ are determined from A-17: $u_1=214.07kJ/kg, \ \alpha_r_1=621.2$ .

The relative specific volume at $s_2$ is calculated from the compression ratio:

$\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825$

#from this, the temperature and enthalpy at state 2, $s_2$ can be determined using interpolations $T_2=862K$ and $h_2=890.9kJ/kg$ . The specific volume at $s_1$ can then be determined as:

$\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg$

Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

$P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa$

.The thermal efficiency=> maximum temperature at $s_3$ can be obtained from the expansion work at constant pressure during $s_2-s_3$

$\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K$

b.Relative SV and enthalpy at $s_3$ are obtained for the given temperature with interpolation with data from A-17 : $a_r_3=4.553 \ and\ h_3=1909.62kJ/kg$

Relative SV at $s_4$ is

$a_r_4=\frac{r}{r_c}\alpha _r_3$

= $=\frac{16}{2}\times4.533\\=36.424$

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

$n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563$

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

$MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa$

Hence, the mean effective relative pressure is 674.95kPa

3 0

3 years ago

Find the frequency of a wave produced by a generator that emits 100 pulses in 2.0 s.

kobusy [5.1K]

The frequency of the wave is 50 Hz

Explanation:

The frequency of the wave is defined as the number of cycles per second of the wave:

$f=\frac{N}{t}$

where

N is the number of cycles completed in a time t.

Frequency is measured in Hertz (Hz).

In this problems, the wave has

N = 100 pulses

in

t = 2.0 s

Therefore, its frequency is

$f=\frac{100}{2.0}=50 Hz$

Learn more about waves and frequency here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

7 0

2 years ago

This time particle A starts from rest and accelerates to the right at 65.5 cm/s

FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

$d = v_i t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}(65.5)t^2$

$d_1 = 32.75 t^2 cm$

Now we know that B moves with constant speed so in the same time B will move to another distance

$d_2 = 44 \times t$

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

$d_1 = 349 + d_2$

$32.75 t^2 = 349 + 44 t$

on solving above kinematics equation we have

$t = 4 s$

4 0

2 years ago

The mechanical energy of a coconut falling from a tree______

Alexxandr [17]

Answer:

D. is the answer

Explanation:

8 0

3 years ago

Read 2 more answers

Which statement shows how to correctly convert from the mass of a compound in grams to the amount of that compound in moles?

expeople1 [14]

Answer:

mass (grams) x 1 =

molar mass (g/mol)

amount (moles)

Explanation:

6 0

3 years ago