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pentagon [3]
3 years ago
7

What would happen if you put a material between each marble? Newton's Cradle

Physics
1 answer:
slamgirl [31]3 years ago
5 0

Answer:

Explanation:

Momentum is the mass of a moving body times its velocity. It can be transferred from one object to another

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Atoms in a magnet are arranged according to their magnetic poles. Their north and south poles determine how the atoms in the magnet are arranged.
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Help me do the c.....​
frez [133]

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as one vector has 2 things which is used to define it ->

  1. Magnitude
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so if both r equal then both vectors r equal vectors

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c)Is \:  equal  \: to

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An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 75
katen-ka-za [31]

Answer:

Part A) 3899 kPa  

Part B) 392.33 kJ/kg  

Part C) 0.523

Part D) 495 kPa

Explanation:

Part A

First from the temperature at state 1 the relative specific volume and the internal energy at that state are determined from:

u_{1} = 214.07 kJ/kg  

\alphar_{1} = 621.2  

The relative specific volume at state 2 is obtained from the compression ratio:  

\alphar_{2} = \frac{\alpha r_{1}  }{r}

     =621.2/ 8

    = 77.65  

From this the temperature and internal energy at state 2 can be determined using interpolation with data from A-17(table):  

T_{2} = 673 K

u_{2} = 491.2 kJ/kg  

The pressure at state 2 can be determined by manipulating the ideal gas relations at state 1 and 2:  

P_{2} =  P_{1} r\frac{T_{2} }{T_{1} }

       = 95*8*673/300

      = 1705 kPa  

Now from the energy balance for stage 2-3 the internal energy at state 3 can be obtained:  

deltau_{2-3} =q_{in}\\ u_{3} -u_{2} =q_{in}\\u_{3}=u_{2}+q_{in}

     = 1241.2 kJ/kg

From this the temperature and relative specific volume at state 3 can be determined by interpolation with data from A-17(table):  

T_{3} = 1539 K  

\alpha r_{3} = 6.588  

The pressure at state 3 can be obtained by manipulating the ideal gas relations for state 2 and 3:  

P_{3} =P_{2} \frac{T_{3} }{T_{2} }

     = 3899 kPa  

<u>Part B</u>

The relative specific volume at state 4 is obtained from the compression ratio:  

\alpha r_{4}= r\alpha r_{3}

      = 52.7

From this the temperature and internal energy at state 4 can be determined by interpolation with data from A-17:  

T_{4}=775 K

u_{4}= 571.74 kJ/kg  

The net work output is the difference of the heat input and heat rejection where the heat rejection is determined from the decrease in internal energy in stage 4-1:  

w=q_{in}-q_{out}\\q_{in}-(u_{4} -u_{1} )\\=392.33 kJ/kg

<u>Part C  </u>

The thermal efficiency is obtained from the work and the heat input:  

η=\frac{w}{q_{in} }

=0.523

<u>Part D  </u>

The mean effective pressure is determined from its standard relation:  

MEP=\frac{w}{\alpha_{1}- \alpha_{2} }

      =\frac{w}{\alpha_{1}(1- \frac{1}{r}  }

      =\frac{rwP_{1} }{RT_{1} (r-1) }

      =495 kPa

8 0
4 years ago
How do you find the oscillation period in seconds for different pendulum lengths?
GalinKa [24]

To find:

The equation to find the period of oscillation.

Explanation:

The period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity.

Thus the period of a pendulum is given by the equation,

T=2\pi\sqrt{\frac{L}{g}}

Where L is the length of the pendulum and g is the acceleration due to gravity.

On substituting the values of the length of the pendulum and the acceleration due to gravity at the point where the period of the pendulum is being measured, the above equation yields the value of the period of the pendulum.

Final answer:

The period of oscillation of a pendulum can be calculated using the equation,

T=2\pi\sqrt{\frac{L}{g}}

3 0
1 year ago
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