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Explain why electricity transmission cables are left sagging and calculate the height of a mountain given the pressure difference in cm of Hg.

Physics

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Answer

1632 m\text{1632 m}

4.5 cm3\text{4.5 cm}^3

Step-by-step solution

4 steps
  1. Answer Question 1.

    Electricity transmission cables are left sagging between pylons to allow for thermal expansion and contraction. This prevents the cables from becoming too taut and snapping during cold weather when they contract, or sagging excessively and touching the ground during hot weather when they expand.

  2. Answer Question 2.

    The pressure difference between the base and the top of the mountain can be expressed in terms of the mercury column and the air column.
    Given:
    Pressure at base, P1=75.0 cm of HgP_1 = 75.0 \text{ cm of Hg}
    Pressure at top, P2=60.0 cm of HgP_2 = 60.0 \text{ cm of Hg}
    Density of air, ρair=1.25 kg/m3\rho_{air} = 1.25 \text{ kg/m}^3
    Density of mercury, ρHg=13600 kg/m3\rho_{Hg} = 13600 \text{ kg/m}^3
    Acceleration due to gravity, g≈9.81 m/s2g \approx 9.81 \text{ m/s}^2 (or 10 m/s210 \text{ m/s}^2 if not specified, but it will cancel out here).

    The pressure difference in terms of mercury column is:

    ΔP=(P1−P2)=(75.0−60.0) cm of Hg=15.0 cm of Hg\Delta P = (P_1 - P_2) = (75.0 - 60.0) \text{ cm of Hg} = 15.0 \text{ cm of Hg}

    Convert this height to meters:

    ΔhHg=15.0 cm=0.15 m\Delta h_{Hg} = 15.0 \text{ cm} = 0.15 \text{ m}

    The pressure difference can also be expressed as ΔP=ρgh\Delta P = \rho g h.
    So, the pressure difference due to the mercury column is:

    ΔP=ρHggΔhHg\Delta P = \rho_{Hg} g \Delta h_{Hg}

    The pressure difference due to the height of the mountain (air column) is:

    ΔP=ρairghmountain\Delta P = \rho_{air} g h_{mountain}

    Equating the two expressions for ΔP\Delta P:

    ρHggΔhHg=ρairghmountain\rho_{Hg} g \Delta h_{Hg} = \rho_{air} g h_{mountain}

    We can cancel gg from both sides:

    ρHgΔhHg=ρairhmountain\rho_{Hg} \Delta h_{Hg} = \rho_{air} h_{mountain}

    Now, solve for hmountainh_{mountain}:

    hmountain=ρHgΔhHgρairh_{mountain} = \frac{\rho_{Hg} \Delta h_{Hg}}{\rho_{air}}

    Substitute the given values:

    hmountain=(13600 kg/m3)(0.15 m)1.25 kg/m3h_{mountain} = \frac{(13600 \text{ kg/m}^3)(0.15 \text{ m})}{1.25 \text{ kg/m}^3} hmountain=20401.25 mh_{mountain} = \frac{2040}{1.25} \text{ m} hmountain=1632 mh_{mountain} = 1632 \text{ m}

    The height of the mountain is 1632 m\boxed{\text{1632 m}}.

  3. Answer Question 3.

    Two factors that would raise the boiling point of water are:

    • Increasing the external pressure on the water.
    • Adding impurities (solutes) to the water.
  4. Answer Question 4.

    Given:
    Initial volume of water in burette = 2.5 cm32.5 \text{ cm}^3
    Number of drops added = 40
    Volume per drop = 0.05 cm30.05 \text{ cm}^3

    First, calculate the total volume of the added drops:

    Total volume of drops=Number of drops×Volume per drop\text{Total volume of drops} = \text{Number of drops} \times \text{Volume per drop} Total volume of drops=40×0.05 cm3\text{Total volume of drops} = 40 \times 0.05 \text{ cm}^3 Total volume of drops=2.0 cm3\text{Total volume of drops} = 2.0 \text{ cm}^3

    Now, calculate the new reading in the burette by adding the initial volume and the total volume of drops:

    New reading=Initial volume+Total volume of drops\text{New reading} = \text{Initial volume} + \text{Total volume of drops} New reading=2.5 cm3+2.0 cm3\text{New reading} = 2.5 \text{ cm}^3 + 2.0 \text{ cm}^3 New reading=4.5 cm3\text{New reading} = 4.5 \text{ cm}^3

    The new reading would be 4.5 cm3\boxed{\text{4.5 cm}^3}.

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