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3 years ago

1. You do not need to remove the lead weights inside tires before recycling them.

Engineering

1 answer:

makvit [3.9K]3 years ago

7 0

It true -.-.-.-.-.-.-.-.-.-.-.-.-.- EZ

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3 years ago

Quadrilateral ABCD is a rectangle.<br> If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.

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3 years ago

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3 years ago

A well-insulated rigid tank contains 5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters of

vlabodo [156]

Answer:

$S_{gen} = 18.519\,\frac{kJ}{K}$

Explanation:

Given that rigid tank is a closed system, the following model is constructed after the First Law of Thermodynamics:

$W_{heater} + U_{sys,1} - U_{sys,2} = 0$

$W_{heater} = U_{sys,2} - U_{sys,1}$

$W_{heater} = m\cdot (u_{2}-u_{1})$

The entropy generation inside the rigid tank is determined by appropriate application of the Second Law of Thermodynamics:

$S_{sys,1} - S_{sys,2} + S_{gen} = 0$

$S_{gen} = S_{sys,2} - S_{sys,1}$

$S_{gen} = m\cdot (s_{2}-s_{1})$

The properties of the steam are obtained from steam tables:

Intial State

$P = 200\,kPa$

$T = 120.21\,^{\textdegree}C$

$\nu = 0.2222\,\frac{m^{3}}{kg}$

$u = 1010.7\,\frac{kJ}{kg}$

$s = 2.9294\,\frac{kJ}{kg\cdot K}$

$x = 0.25$

Final State

$P = 869.567\,kPa$

$T = 173.88\,^{\textdegree} C$

$\nu = 0.2222\,\frac{m^{3}}{kg}$

$u = 2578.6\,\frac{kJ}{kg}$

$s = 6.6332\,\frac{kJ}{kg\cdot K}$

$x = 1.00$

The entropy change of the steam during the process is:

$S_{gen} = (5\,kg)\cdot \left(6.6332\,\frac{kJ}{kg\cdot K} - 2.9294\,\frac{kJ}{kg\cdot K} \right)$

$S_{gen} = 18.519\,\frac{kJ}{K}$

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4 years ago

What safety mechanism is sometimes disabled on a nail gun leading you to

Irina-Kira [14]

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Hoped this Helped!!!!!

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4 years ago