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3 years ago

Which galaxy is the most stretched out?

Physics

2 answers:

azamat3 years ago

5 0

The answer is the oddball spiral galaxy. hope this help

RideAnS [48]3 years ago

4 0

Answer:

this is the anwser

Explanation:

The oddball spiral galaxy, called Messier 66, is one-thirdof the Leo Triplet, a group of three interacting galaxies about 35 millionlight-years from Earth (a light-year is the distance light can cover in ayear).

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An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

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<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

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<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<em>Ideal gas is allowed to expand isothermally:</em>

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

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<em>Next we will calculate the work done on the gas:</em>

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

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<em>Using the same method as above:</em>

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

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<em>Next we will calculate the work done on the gas:</em>

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

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<em>Finally we could calculate the total work done and heat added as follows:</em>

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

$\texttt{ }$

<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

3 years ago

A baseball with a mass of 0.80 kg is given an acceleration of 20.00 m/s. How much net force was applied to the ball

dybincka [34]

a x m = f

.80 x 20 = 16

4 0

3 years ago

Which process is an example of a physical change?

Anit [1.1K]

Answer:

Option A

Carrots are cut into small pieces and mixed into a salad

Explanation:

When physical changes occur, the actual composition remain the same but the molecules are re-arranged. Therefore, when carrots are cut into smaller pieces and mixed into salad, there will be no chemical reaction hence the actual composition will remain the same despite being cut and molecules in it re-arranged. Considering the other options, new substances are formed hence they are deemed as chemical changes. Therefore, option A is correct.

8 0

2 years ago

A jetliner is traveling east from Salt Lake City to Washington DC, a distance of 1,850 miles. The jetliner travels at an average

maks197457 [2]

The jetliner is traveling against the wind. The net speed of the jetliner is
590 mph - 36 mph = 554 mph
The time it takes for the jetliner to arrive at the destination is
1850 miles / 554 mph = 3.34 hours

3 0

3 years ago

The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.17 Hz, and the acc

aksik [14]

Answer:

d = 8.4 cm

Explanation:

In order to calculate the amplitude of oscillation of the top of the building, you use the following formula for the max acceleration of as simple harmonic motion:

$a_{max}=A\omega^2$ (1)

A: amplitude of the oscillation

w: angular speed of the oscillation = 2 $\pi$ f

f: frequency = 0.17Hz

The maximum acceleration of the top of the building is a 2.0% of the free-fall acceleration. Then, you have:

$a_{max}=0.02(9.8m/s^2)=0.196\frac{m}{s^2}$

Then, you solve for A in the equation (1) and replace the values of the parameters:

$A=\frac{a_{max}}{\omega^2}=\frac{a_{max}}{4\p^2i f^2}\\\\A=\frac{0.196m/s^2}{16\pi^2(0.17Hz)^2}\\\\A=0.042m=4.2cm$

The total distance, side to side, of the oscilation of the top of the building is twice the amplitude A. Then you obtain:

d = 2A = 2(4.2cm) = 8.4cm

4 0

3 years ago

Which galaxy is the most stretched out?​

Which galaxy is the most stretched out?