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nika2105 [10]
3 years ago
11

For the MgO–Al2O3 system, what is the maximum temperature that is possible without the formation of a liquid phase? At what comp

osition or over what range of compositions will this maximum temperature be achieved?
Engineering
1 answer:
blagie [28]3 years ago
5 0

Answer:

approximately 2800

At what composition or over what range of compositions will this maximum temperature be achieved? Answer This problem asks that we specify, for the MgO-Al2O3system, Figure 12.23, the maximum temperature without the formation of a liquid phase; it is approximately 2800C which is possible for pure MgO.

Explanation:

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A closely wound, circular coil with radius 2.80 cm has 720 turns. Part APart complete What must the current in the coil be if th
lara [203]

Answer:

I = 4.642 Ampere.

x = 2.145 cm

Explanation:

a) As we know, the magnetic field on the axis of the loop is given as

   B = \frac{ \mu NIa^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} }   }

where  a = radius of loop        

            x = point on the axis of loop

            N = No of turns of coil

Current in the loop for which the magnetic field at the center is 0.0750 Tesla is given as x = 0

Therefore, the above equation can be rewritten as

               B_{x}  = \frac{ \mu NI}{2a}

               I = \frac{2aB}{μN}

by putting values

I =\frac{2 X 0.0750 T X 0.028 m}{4\pi X 10^{-7} T.\frac{m}{A}  X 720}

Current in the loop for which the magnetic field at the center is 0.0750 Tesla  = I = 4.642 Ampere

b)

      Now for part b the magnetic field at a distance x from the center is given as

         B = \frac{ \mu N I a^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} }  }

multiply and divide by a on both sides we get

     B = \frac{ \mu NI}{2a} X \frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} }   }

   As we know, according to Biot sivorts law,the Magnetic Field at the Center of a circular loop is given as

   B = \frac{ \mu NI}{2a}  ( Magnetic field at center) = B_{c}

So we got magnetic field at any point x as

B_{x} = B_{c} X  \frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} }   }

For magnetic field at x is half of the B at center

          B_{x} = \frac{1}{2} B_{c}

from the above two equations

\frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} }   } = \frac{1}{2}

   (x^{2} + a^{2})^{3} = 4a^{6}

   x = \sqrt{4^{\frac{1}{3}} -1 }  X a^{} 

Putting a = 2.80 cm

We have  x = 2.145 cm Ans

           

μ

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Explanation:

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80kg of NaOH is required to neutralize 98kg of sulfuric acid

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Explanation:

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Answer:

Answer to both the parts A and B are provided in the word document attached.

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