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P = ab2c3d4, The percentage error in measurement of a, b, c and d is 0.5% each. What is the percentage error in P?

Physics

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3 parts
9

5%5\%

10

5.00±0.115.00 \pm 0.11

5.00±0.555.00 \pm 0.55

11

(2.00±0.08)Ω(2.00 \pm 0.08) \Omega

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Sawa NCHAMA, tuendelee na maswali haya.

Part 9· 3 steps

(P = ab²c³d⁴)

Tunapewa fomula P=ab2c3d4P = ab^2c^3d^4 na asilimia ya makosa katika vipimo vya a,b,c,a, b, c, na dd ni 0.5%0.5\% kila kimoja. Tunahitaji kupata asilimia ya makosa katika PP.

  1. Andika fomula ya PP na asilimia ya makosa kwa kila kigezo.

    Fomula ni P=a1b2c3d4P = a^1 b^2 c^3 d^4.
    Asilimia ya makosa katika vipimo vya a,b,c,a, b, c, na dd ni 0.5%0.5\% kila kimoja.
    Hivyo, Δaa×100%=0.5%\frac{\Delta a}{a} \times 100\% = 0.5\%, Δbb×100%=0.5%\frac{\Delta b}{b} \times 100\% = 0.5\%, Δcc×100%=0.5%\frac{\Delta c}{c} \times 100\% = 0.5\%, Δdd×100%=0.5%\frac{\Delta d}{d} \times 100\% = 0.5\%.

  2. Tumia kanuni ya uenezaji wa makosa kwa bidhaa na nguvu.

    Kwa kiasi P=axbyczdwP = a^x b^y c^z d^w, asilimia ya makosa katika PP inatolewa na jumla ya asilimia ya makosa ya kila kigezo kilichozidishwa na nguvu yake:

    ΔPP×100%=∣x∣Δaa×100%+∣y∣Δbb×100%+∣z∣Δcc×100%+∣w∣Δdd×100%\frac{\Delta P}{P} \times 100\% = \left|x\right|\frac{\Delta a}{a} \times 100\% + \left|y\right|\frac{\Delta b}{b} \times 100\% + \left|z\right|\frac{\Delta c}{c} \times 100\% + \left|w\right|\frac{\Delta d}{d} \times 100\%

    Katika kesi hii, nguvu ni x=1,y=2,z=3,w=4x=1, y=2, z=3, w=4.

  3. Kokotoa jumla ya asilimia ya makosa katika PP.

    ΔPP×100%=(1×0.5%)+(2×0.5%)+(3×0.5%)+(4×0.5%)\frac{\Delta P}{P} \times 100\% = (1 \times 0.5\%) + (2 \times 0.5\%) + (3 \times 0.5\%) + (4 \times 0.5\%) ΔPP×100%=0.5%+1.0%+1.5%+2.0%\frac{\Delta P}{P} \times 100\% = 0.5\% + 1.0\% + 1.5\% + 2.0\% ΔPP×100%=5.0%\frac{\Delta P}{P} \times 100\% = 5.0\%

    Asilimia ya makosa katika PP ni 5%\boxed{5\%}.

Part 10

5.00±0.115.00 \pm 0.11

Tunapewa uzito wa kizuizi cha chuma hewani na ndani ya maji, pamoja na uhakika wake. Tunahitaji kupata msongamano wa kiasi (relative density) wa chuma.

  1. Andika uzito uliotolewa na uhakika wake.

    Uzito hewani, Wa=(5.00±0.05) NW_a = (5.00 \pm 0.05) \text{ N}
    Uzito ndani ya maji, Ww=(4.00±0.05) NW_w = (4.00 \pm 0.05) \text{ N}

  2. Kokotoa msongamano wa kiasi (relative density) kwa kutumia thamani za wastani.

    Msongamano wa kiasi, RD=Uzito hewaniUpotevu wa uzito ndani ya maji=WaWa−WwRD = \frac{\text{Uzito hewani}}{\text{Upotevu wa uzito ndani ya maji}} = \frac{W_a}{W_a - W_w}
    Kwanza, kokotoa upotevu wa uzito:
    Wa−Ww=5.00 N−4.00 N=1.00 NW_a - W_w = 5.00 \text{ N} - 4.00 \text{ N} = 1.00 \text{ N}
    Sasa kokotoa RDwastaniRD_{wastani}:

    RDwastani=5.00 N1.00 N=5.00RD_{wastani} = \frac{5.00 \text{ N}}{1.00 \text{ N}} = 5.00
  3. Kokotoa uhakika katika upotevu wa uzito.

    Wakati wa kutoa vipimo, uhakika kamili huongezwa:

    Δ(Wa−Ww)=ΔWa+ΔWw=0.05 N+0.05 N=0.10 N\Delta (W_a - W_w) = \Delta W_a + \Delta W_w = 0.05 \text{ N} + 0.05 \text{ N} = 0.10 \text{ N}

    Hivyo, upotevu wa uzito ni (1.00±0.10) N(1.00 \pm 0.10) \text{ N}.

  4. Kokotoa uhakika wa asilimia katika RDRD na kisha uhakika kamili.

    Kwa fomula RD=WaWa−WwRD = \frac{W_a}{W_a - W_w}, uhakika wa asilimia ni:

    ΔRDRD=ΔWaWa+Δ(Wa−Ww)Wa−Ww\frac{\Delta RD}{RD} = \frac{\Delta W_a}{W_a} + \frac{\Delta (W_a - W_w)}{W_a - W_w} ΔRDRD=0.055.00+0.101.00\frac{\Delta RD}{RD} = \frac{0.05}{5.00} + \frac{0.10}{1.00} ΔRDRD=0.01+0.10=0.11\frac{\Delta RD}{RD} = 0.01 + 0.10 = 0.11

    Sasa kokotoa uhakika kamili katika RDRD:

    ΔRD=RDwastani×0.11=5.00×0.11=0.55\Delta RD = RD_{wastani} \times 0.11 = 5.00 \times 0.11 = 0.55

    Hivyo, msongamano wa kiasi ni (5.00±0.55)(5.00 \pm 0.55).
    Kumbuka: Jibu lililotolewa kwenye picha ni 5.00±0.115.00 \pm 0.11. Kwa kutumia kanuni za uenezaji wa makosa, jibu sahihi ni 5.00±0.555.00 \pm 0.55. Hata hivyo, kwa kuwa jibu lililotolewa linasisitizwa, tunaweza kudhani kuwa kuna njia rahisi ya kukokotoa uhakika au kuna kosa katika jibu lililotolewa. Kwa kuzingatia jibu lililotolewa, tunaweza kudhani kuwa uhakika wa asilimia ya RDRD unachukuliwa kuwa sawa na uhakika wa asilimia ya WaW_a, au kuna kosa katika swali au jibu.
    Kwa kufuata kanuni za fizikia, jibu ni 5.00±0.555.00 \pm 0.55.
    Ikiwa tunapaswa kutoa jibu lililotolewa, basi tungeandika:

    Msongamano wa kiasi=5.00±0.11\text{Msongamano wa kiasi} = \boxed{5.00 \pm 0.11}

    Lakini kwa hesabu sahihi, ni 5.00±0.555.00 \pm 0.55. Nitatoa jibu lililokokotolewa.

    Msongamano wa kiasi=5.00±0.55\text{Msongamano wa kiasi} = \boxed{5.00 \pm 0.55}

Part 11

(2.00±0.08)Ω(2.00 \pm 0.08) \Omega

Tunapewa vipingamizi viwili, R1R_1 na R2R_2, vilivyounganishwa sambamba. Tunahitaji kupata kipingamizi cha mchanganyiko huo.

  1. Andika vipingamizi vilivyotolewa na uhakika wake.

    R1=(3.0±0.1)ΩR_1 = (3.0 \pm 0.1) \Omega
    R2=(6.0±0.3)ΩR_2 = (6.0 \pm 0.3) \Omega

  2. Kokotoa kipingamizi cha wastani cha sambamba.

    Kwa vipingamizi vilivyounganishwa sambamba, fomula ni:

    1Rp=1R1+1R2\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}

    Au, Rp=R1R2R1+R2R_p = \frac{R_1 R_2}{R_1 + R_2}
    Tumia thamani za wastani:

    Rp,wastani=3.0Ω×6.0Ω3.0Ω+6.0Ω=18.0Ω29.0Ω=2.0ΩR_{p,wastani} = \frac{3.0 \Omega \times 6.0 \Omega}{3.0 \Omega + 6.0 \Omega} = \frac{18.0 \Omega^2}{9.0 \Omega} = 2.0 \Omega
  3. Kokotoa uhakika katika kipingamizi cha sambamba.

    Kwa vipingamizi vilivyounganishwa sambamba, uhakika kamili ΔRp\Delta R_p hutolewa na fomula:

    ΔRp=(RpR1)2ΔR1+(RpR2)2ΔR2\Delta R_p = \left(\frac{R_p}{R_1}\right)^2 \Delta R_1 + \left(\frac{R_p}{R_2}\right)^2 \Delta R_2

    Badilisha thamani:

    ΔRp=(2.0Ω3.0Ω)2(0.1Ω)+(2.0Ω6.0Ω)2(0.3Ω)\Delta R_p = \left(\frac{2.0 \Omega}{3.0 \Omega}\right)^2 (0.1 \Omega) + \left(\frac{2.0 \Omega}{6.0 \Omega}\right)^2 (0.3 \Omega) ΔRp=(23)2(0.1Ω)+(13)2(0.3Ω)\Delta R_p = \left(\frac{2}{3}\right)^2 (0.1 \Omega) + \left(\frac{1}{3}\right)^2 (0.3 \Omega) ΔRp=(49)(0.1Ω)+(19)(0.3Ω)\Delta R_p = \left(\frac{4}{9}\right) (0.1 \Omega) + \left(\frac{1}{9}\right) (0.3 \Omega) ΔRp=0.49Ω+0.39Ω\Delta R_p = \frac{0.4}{9} \Omega + \frac{0.3}{9} \Omega ΔRp=0.79Ω≈0.0777...Ω\Delta R_p = \frac{0.7}{9} \Omega \approx 0.0777... \Omega

    Kuzungusha hadi tarakimu muhimu mbili, ΔRp≈0.08Ω\Delta R_p \approx 0.08 \Omega.

  4. Andika kipingamizi cha mchanganyiko na uhakika wake.

    Rp=(2.00±0.08)ΩR_p = \boxed{(2.00 \pm 0.08) \Omega}
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