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P varies as the square of R. R varies as the square of T. When P=18, R=3 and T=5. Express P in terms of T hence find P when T=10.

Mathematics

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Answer

288\text{288}

Step-by-step solution

5 steps

Huu hapa ni ufumbuzi wa swali la 10:

  1. Andika uhusiano kati ya P na R.

    P inatofautiana moja kwa moja na mraba wa R. Hii inaweza kuandikwa kama:

    P=k1R2P = k_1 R^2

    Ambapo k1k_1 ni kigezo cha uwiano.

  2. Andika uhusiano kati ya R na T.

    R inatofautiana moja kwa moja na mraba wa T. Hii inaweza kuandikwa kama:

    R=k2T2R = k_2 T^2

    Ambapo k2k_2 ni kigezo cha uwiano.

  3. Unganisha uhusiano ili kueleza P kwa kutumia T.

    Badilisha RR kutoka Hatua ya 2 kwenye mlinganyo wa Hatua ya 1:

    P=k1(k2T2)2P = k_1 (k_2 T^2)^2 P=k1k22T4P = k_1 k_2^2 T^4

    Tunaweza kuweka K=k1k22K = k_1 k_2^2, hivyo mlinganyo unakuwa:

    P=KT4P = K T^4
  4. Tafuta thamani ya vigezo k1k_1 na k2k_2 (au K) kwa kutumia thamani zilizopewa.

    Tunapewa P=18P = 18, R=3R = 3, na T=5T = 5.
    Kutoka P=k1R2P = k_1 R^2:

    18=k1(3)218 = k_1 (3)^2 18=9k118 = 9k_1 k1=189=2k_1 = \frac{18}{9} = 2

    Kutoka R=k2T2R = k_2 T^2:

    3=k2(5)23 = k_2 (5)^2 3=25k23 = 25k_2 k2=325k_2 = \frac{3}{25}

    Sasa, badilisha k1k_1 na k2k_2 kwenye mlinganyo wa P kwa kutumia T:

    P=k1k22T4P = k_1 k_2^2 T^4 P=2(325)2T4P = 2 \left(\frac{3}{25}\right)^2 T^4 P=2(9625)T4P = 2 \left(\frac{9}{625}\right) T^4 P=18625T4P = \frac{18}{625} T^4

    Huu ndio usemi wa P kwa kutumia T.

  5. Tafuta P wakati T = 10.

    Tumia mlinganyo uliopatikana katika Hatua ya 4:

    P=18625(10)4P = \frac{18}{625} (10)^4 P=18625(10000)P = \frac{18}{625} (10000) P=18×10000625P = 18 \times \frac{10000}{625} P=18×16P = 18 \times 16 P=288P = 288

    Thamani ya P wakati T = 10 ni 288\boxed{\text{288}}.

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