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Find g if E = 1200 Joules, m = 18.5 kg, v = 10 m/s, h = 1.5 m. And make L the subject.

Physics

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Answer

g=9.91 m/s2g = \textbf{9.91 m/s}^2

L=gT24π2L = \frac{gT^2}{4\pi^2}

Step-by-step solution

7 steps

b) To find gg, we use the formula derived in the previous step: g=2E−MV22Mhg = \frac{2E - MV^2}{2Mh}.
Given values:
E=1200E = 1200 Joules
M=18.5M = 18.5 kg
V=10V = 10 m/s
h=1.5h = 1.5 m

  1. Substitute the given values into the formula.

    g=2(1200)−(18.5)(10)22(18.5)(1.5)g = \frac{2(1200) - (18.5)(10)^2}{2(18.5)(1.5)}

  2. Calculate the numerator.

    2(1200)−(18.5)(10)2=2400−(18.5)(100)2(1200) - (18.5)(10)^2 = 2400 - (18.5)(100) =2400−1850=550= 2400 - 1850 = 550

  3. Calculate the denominator.

    2(18.5)(1.5)=37×1.5=55.52(18.5)(1.5) = 37 \times 1.5 = 55.5

  4. Divide the numerator by the denominator to find gg.

    g=55055.5≈9.9099g = \frac{550}{55.5} \approx 9.9099

    Rounding to three significant figures:

    g=9.91 m/s2\boxed{g = \textbf{9.91 m/s}^2}

    2) To make LL the subject of the formula T=2πLgT = 2\pi \sqrt{\frac{L}{g}}:

  5. Divide both sides by 2π2\pi to isolate the square root term.

    T2π=Lg\frac{T}{2\pi} = \sqrt{\frac{L}{g}}

  6. Square both sides of the equation to eliminate the square root.

    (T2π)2=(Lg)2\left(\frac{T}{2\pi}\right)^2 = \left(\sqrt{\frac{L}{g}}\right)^2 T2(2π)2=Lg\frac{T^2}{(2\pi)^2} = \frac{L}{g} T24π2=Lg\frac{T^2}{4\pi^2} = \frac{L}{g}

  7. Multiply both sides by gg to make LL the subject.

    L=g×T24π2L = g \times \frac{T^2}{4\pi^2} L=gT24π2\boxed{L = \frac{gT^2}{4\pi^2}}
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