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ICE Princess25 [194]
3 years ago
12

Help I need an answer!

Physics
1 answer:
Umnica [9.8K]3 years ago
5 0

You apparently haven't noticed yet . . .

The moon rises in the East, moves across the sky, and sets
in the West, just like the sun and everything else in the sky.

It takes the moon about 12 hours and 25 minutes to go from
rising to setting, and then it rises again about 24 hours and
49 minutes later.  So each day, the moon rises about 49 minutes
later than it rose on the previous day.

In order to describe the direction and height of the moon in the sky,
you need to tell the date and the time, and also the latitude on Earth
from which you observed it.  And wherever the moon was in the sky
at that time, it was in a noticeably different place 30 minutes later.

Just like the sun.

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A industrial (large) pressure cooker operates at 275 kPa. Initially there is 10 kg of water at 20°C, the cooker is operated unt
alex41 [277]

Answer:

Q_{in} = 25349.92\,kJ

Explanation:

Let establish a control volume in the industrial pressure cooker, which is a transient state system. From the First Law of Thermodynamics, the heating process is modelled:

Q_{in} + m_{1}\cdot h_{1} - m_{2}\cdot h_{2} = (m_{1}-m_{2})\cdot u_{2} - m_{1}\cdot u_{1}

The heat transfered to the cooker is:

Q_{in} = m_{2}\cdot h_{2} - m_{1}\cdot h_{1} + (m_{1}-m_{2})\cdot u_{2}-m_{1}\cdot u_{1}

Properties at each state are described below:

State 1

u_{1} = 83.913\,\frac{kJ}{kg}

h_{1} = 83.915\,\frac{kJ}{kg}

State 2

u_{2} = 2540.1\,\frac{kJ}{kg}

h_{2} = 2720.9\,\frac{kJ}{kg}

The heat transfered to the cooker is:

Q_{in} = (9\,kg)\cdot (2720.9\,\frac{kJ}{kg} ) - (10\,kg)\cdot (83.915\,\frac{kJ}{kg} ) + (1\,kg) \cdot (2540.1\,\frac{kJ}{kg} )-(10\,kg)\cdot (83.913\,\frac{kJ}{kg} )

Q_{in} = 25349.92\,kJ

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In a thermally isolated environment, you add ice at 0°C and steam at 100°C. (a) Determine the amount of steam condensed (in g) a
kondor19780726 [428]

Answer:

 a) 11.2 g

   b) 3.73 g.

Explanation:

a) If we assume temperature of mixture to be 100°C , heat released by steam will be 11.2 x 540 = 6048 cals and heat gain gained by will be

79 x 80 + 79 x 1 x 100 = 14220 cals . Since former heat is less than later heat ,water will not be warmed up to 100°C. Let equilibrium temperature be t .  

Heat gained by water = 79 x 80 + 79 x 1 x t = 11.2 x 540 + 11.2( 100 - t )

t = 9.4°

amount of steam condensed = 11.2 g.

b) In this case, whole of water will be warmed up to 100°C as steam is much .heat required by water to warm up to boiling point

= 11.2 x 80 + 11.2 x 100 = 2016 cals

amount of steam condensed =  2016 / 540 = 3.73 g .  

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