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arsen [322]
3 years ago
6

1 kg air in a piston-cylinder assembly is heated at constant pressure, resulting the expansion of the volume. the initial temper

ature of the air was 300 k, and the air temperature becomes 500 k after the expansion. what is the boundary work done by the air

Physics
1 answer:
rewona [7]3 years ago
4 0

Answer:

57,42 KJ

Explanation:

By a isobaric proces, the expresion for the works in the jpg adjunt. Then:

W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)

By the ideal gases law: PV=RTn

Then, in (1): (remember Pa = Pb)

W =  R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)

Since we have 1 Kg air: How much is this in moles?

From bibliography: 28.96 g/mol

Then, in 1 Kg (1000 g) there are:

n =  34,53 mol

Finally, in (2):

W =  (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ

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The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th
worty [1.4K]

Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg

where m_a is the mass of the arrow, m_b is the mass of the block, \Delta H of the change in height of the block after the collision, and v is the velocity of the arrow before it hit the block.

Solving for the velocity v, we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} } $

and we put in the numerical values

m_a = 0.045kg,

m_b = 1.40kg,

\Delta H = 0.4m,

g= 9.8m/s^2

and simplify to get:

\boxed{ v= 15.9m/s}

The arrow was moving at 15.9 m/s

6 0
3 years ago
What magnification would be obtained if an eyepiece with a focal length of 0.38 m was placed on telescope?
weqwewe [10]

Answer:

This question is incomplete

Explanation:

This question is incomplete because the telescope's focal length was not provided. The formula to be used here is

Magnification = telescope's focal length/eyepiece's focal length

The eyepiece's focal length was provided in the question as 0.38 m.

NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).

7 0
2 years ago
Sometimes balance point may not be obtained on the potentiometer wire why​
scoundrel [369]
Solution
Let a cell of emf E be connected across the entire length L of a potentiometer wire . Now , if the balance point is obtained at a length l during measurement of an unknown voltage

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V > E
8 0
2 years ago
Use what you know about reflection and absorption of light to complete the sentence.
Blababa [14]

A red ladybug appears red in white light, red in red light, and black in blue light. Those would be the proper selections you'd need.

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3 years ago
Read 2 more answers
Laskar, J.: 1990, The chaotic motion of the solar system. A numerical estimate of the size of the chaotic zones, Icarus, 88, 266
balandron [24]

The chaotic nature of the Solar System excluding Pluto was established by the numerical computation of the maximum Lyapunov exponent of its secular system over 200 myr.

<h3>What is chaotic motion of the solar system ?</h3>

There has been an increase in awareness of chaotic dynamics in the solar system over the past 20 years. The orbits of tiny objects in the solar system, such as asteroids, comets, and interplanetary dust, are now known to be chaotic and to experience significant variations across geological time periods.

  • a completely unpredictable orbit, or one where significant changes in the orbit can result from even small changes in the position and/or velocity of the orbiting entity.

Learn more about Chaotic motion here:

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1 year ago
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