What are the two outcomes for a proto star?

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

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in

wel

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:

Potential energy = PE = ?

Work done = W = ?

Answer:

Potential energy = 0.343 Joules

Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

3 0

3 years ago

Suppose that Lisa walks her dog around the block for a little exercise. The block is 1 mile around. If she walks around the bloc

natita [175]

Displacement only measure how far between the starting and ending point. In this case, Lisa walks around the block as a circle so the starting point is the same as the ending point. Thus, displacement is 0mile.
On the other hand, distance measures exactly how far she walks. In this case, the distance is 1 mile, same as the perimeter of the block.

7 0

3 years ago

Which statements describe using genetic factors to influence the growth of organisms? Select the three (3) that apply.

Damm [24]

Answer:

- increasing use of hybrid crops

- altering genes in DNA to create new plants

- developing disease or pest resistant crops

Explanation:

The use of genetic factors to influence the growth of a plant encompasses manipulating the genetic constituent (gene) of such plant.

For example,

- Increasing use of hybrid crops entails mating two pure bred plants based on a gene of interest responsible for a particular trait, to form a hybrid.

- Altering genes in DNA to create new plants is also a genetic factor as it has to with gene modification.

- developing disease or pest resistant crops means that the genetic make up of such plant has been modified to be resistant to pest/disease.

6 0

3 years ago

When you attract every object in the universe with a force that is proportional to the mass of the objects and to the distance b

yuradex [85]

When you attract every object in the universe with a force that is proportional to the mass of the objects and to the distance between them, we are obeying Newton's law of universal gravitation.

<h3>Newton's law of universal gravitation</h3>

Newton's law of universal gravitation states that the force of attraction between two masses in the universe is directly proportional to the product of the masses and inversely proportional to the the square of the distance between them.

The mathematical interpretation of the above law is

F ∝Mm/r²

Removing the proportionality sign,

F = GMm/r².

Where:

F = Force of attraction
G = Gravitational constant
M = Bigger mass
m = Smaller mass
r = Distance between the masses.

From the above, When you attract every object in the universe with a force that is proportional to the mass of the objects and to the distance between them, we are obeying Newton's law of universal gravitation.

Learn more about Newton's law of universal gravitation here: brainly.com/question/9373839

#SPJ12

7 0

2 years ago