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valentinak56 [21]
3 years ago
9

Air is compressed in a cylinder such that the volume changes from 100.0 to 10.0 in^3. The initial pressure is 50.0 psia and the

temperature is held at a constant 100.0 °F. How much work is required to change the volume? (Hint you will need to identify what psia means)
Physics
1 answer:
EleoNora [17]3 years ago
7 0

Answer:

5953.42 J

Explanation:

Given:

Initial volume, V_i= 100 in³

Final Volume, V_f = 10 in³

Initial pressure = 50 psia

Temperature = 100° F = 310.93 K

For isothermal reversible process, work done is given as:

Work done = -2.303RTlog_{10}\frac{V_f}{V_i}

Where,

R is the ideal gas constant = 8.314 J/mol.K

or

Work done = -2.303\times8.314\times310.93log_{10}\frac{10}{100}

or

Work done = 5953.42 J

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Input energy is: 200 joule
Output energy is: 100 joule

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How does the use of renewable sources of energy help to solve the problem of energy
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Answer:

The use of renewable sources of energy help to solve the problem of energy  crisis is discussed below in details.

Explanation:

Environmental and economic advantages of adopting renewable energy incorporate: Creating energy that originates no greenhouse gas discharges from fossil fuels and decreases some kinds of air pollution. Increasing energy stocks and decreasing dependency on shipped fuels.

Here are 5 actionable suggestions you can follow to make a decent utilization of renewable energy on your bases:

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3 years ago
На рычаге размещены два противовеса таким образом, что рычаг находится в состоянии равновесия. Вес расположенного слева противов
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Answer:

Вес противовеса, или сила нагрузки, составляет затем 100 000 фунтов-футов, разделенных на 20 футов, или 5000 фунтов.

Explanation:

8 0
4 years ago
An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
Why does a black hole have a stronger gravitational pull than the star that collapse to form it?​
Studentka2010 [4]

Answer:

We consider Black Holes as an object that possesses extreme gravitational pull, but wait aren’t they have the same mass(or less) as that of their parent star. And we know that gravitational pull ‘F’ is directly proportional to the mass of an object, so if the mass is same(or less) then why do black holes have stronger gravity than the stars they evolved from.

The above consideration that F is directly proportional to the mass is partially correct, one should also mention that F is also inversely proportional to the square of the distance between the considered objects.

F = G*(M*m)/(r^2)

Where:

· F is the force acting on you due to star

· M is the mass of Parent star / Black Hole

· m is the mass of an observer, here it is you

· r is the radial distance between the star and you

We know that black hole formed, has much smaller size than that of its parent star and all that mass is compressed to a much smaller scale. If you consider a Star as having a size of an earth then the black hole formed will have a size of small city.

Let us say that you are standing at an r distance away from a star (r>R1), where R1 is the radius of the star, of course (R1>R2), where R2 is the radius of Black Hole.

The Force by which the star in case 1 attracts you will be equal(or less) to the force by which black hole in case 2. So, there is nothing increase in gravitational pull, it is same(or less) as that of the parent star.

Wait a minute, then why people say that black holes have massive gravitational pull.

The gravitational pull increases as we move closer to the black hole, and when we are at its surface, it is enormous as compare to its star surface, because of the difference in the size.

We know that gravitational pull not only depends upon the mass but also depends upon the radial distance between the concerned objects here, it is you and the black hole.

Here, the size of the black hole is much smaller than that of its parent star, i.e (R1>>>R2), and thus we get F1<<<F2, and that is why we say that the black hole has enormous gravitational pull, such that nothing can escape, not even light.

8 0
3 years ago
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