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djverab [1.8K]
4 years ago
7

Why do magnetic bodies lose their magnetic force?

Physics
1 answer:
stiks02 [169]4 years ago
8 0

Answer:

Well a s the temperature increases, at a certain point called the Curie temperature, a magnet will lose its strength completely. So once the metal cools, its ability to attract magnets returns, though its permanent magnetism becomes weak. In general heat is the factor that has the most effect on permanent magnets.Explanation:

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What happened in my room<br>​
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A cyclist and his bicycle have a combined mass of 85 kg. If he is moving at
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211.2 kg:m/s

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The group number of on the periodic table corresponds to which characteristics of the elements of that group ?
vovangra [49]

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Explanation:

In 1789, Antoine Lavoisier published a list of 33 chemical elements. Although Lavoisier grouped the elements into gases, metals, non-metals, and earths, chemists spent the following century searching for a more precise classification scheme. In 1829, Johann Wolfgang Döbereiner observed that many of the elements could be grouped into triads (groups of three) based on their chemical properties. Lithium, sodium, and potassium, for example, were grouped together as being soft, reactive metals. Döbereiner also observed that, when arranged by atomic weight, the second member of each triad was roughly the average of the first and the third.[19] This became known as the Law of Triads.[20] German chemist Leopold Gmelin worked with this system, and by 1843 he had identified ten triads, three groups of four, and one group of five. Jean Baptiste Dumas published work in 1857 describing relationships between various groups of metals. Although various chemists were able to identify relationships between small groups of elements, they had yet to build one scheme that encompassed them all.[19]

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Two red and two overlapping balls in the center are surrounded by a green, fuzzy, circular cloud with a white line running throu
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5 0
2 years ago
Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un
Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

          K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}

  \gamma = 1,     K_{IC} = 24.1

  [/tex]\sigma_{y}[/tex] = 570

and,   \sigma_{f} = 570 \times \frac{3}{4}

                       = 427.5

Hence, we will calculate the critical crack length as follows.

      a = \frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}

        = \frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}

       = 10.13 \times 10^{-4}

Therefore, largest size is as follows.

            Largest size = 2a

                                 = 2 \times 10.13 \times 10^{-4}

                                 = 20.26 \times 10^{-4}

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is 20.26 \times 10^{-4}.

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