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Distinguish between hay and silage, list 3 protein sources for poultry, and calculate energy and velocity of a falling body.

Physics

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ii

17.15 m/s\text{17.15 m/s}

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here are the answers to the questions:

2(a)(i) Distinguish between hay and silage in livestock production.
Hay is forage (like grass or legumes) that has been cut and dried to reduce its moisture content, preserving it for storage. Silage is forage that has been cut and then fermented in an anaerobic environment, preserving it with a higher moisture content.

2(a)(ii) List three sources of protein in poultry feedstuff.

  • Soybean meal
  • Fish meal
  • Groundnut cake

2(b) A body of mass 10 kg10 \text{ kg} is placed at a height of 20 m20 \text{ m} above the ground. The body was made to fall to a height of 5 m5 \text{ m} above the ground.

Part i

What energy will the body have at height 5 m5 \text{ m} above the ground?

At 5 m5 \text{ m} above the ground, the body will have both potential energy (due to its height) and kinetic energy (due to its motion).

Part ii

17.15 m/s\text{17.15 m/s}

Calculate the velocity of the body at height 5 m5 \text{ m} above the ground.

  1. Identify given values and constants.

    Mass m=10 kgm = 10 \text{ kg}
    Initial height h1=20 mh_1 = 20 \text{ m}
    Final height h2=5 mh_2 = 5 \text{ m}
    Initial velocity v1=0 m/sv_1 = 0 \text{ m/s} (since it was placed)
    Acceleration due to gravity g=9.8 m/s2g = 9.8 \text{ m/s}^2

  2. Apply the principle of conservation of mechanical energy.

    The total mechanical energy at the initial height equals the total mechanical energy at the final height.

    PE1+KE1=PE2+KE2PE_1 + KE_1 = PE_2 + KE_2 mgh1+12mv12=mgh2+12mv22mgh_1 + \frac{1}{2}mv_1^2 = mgh_2 + \frac{1}{2}mv_2^2

    Since v1=0v_1 = 0, the initial kinetic energy is zero.

    mgh1=mgh2+12mv22mgh_1 = mgh_2 + \frac{1}{2}mv_2^2
  3. Solve for the final velocity v2v_2.

    Divide by mm:

    gh1=gh2+12v22gh_1 = gh_2 + \frac{1}{2}v_2^2

    Rearrange to find v22v_2^2:

    12v22=gh1−gh2\frac{1}{2}v_2^2 = gh_1 - gh_2 12v22=g(h1−h2)\frac{1}{2}v_2^2 = g(h_1 - h_2) v22=2g(h1−h2)v_2^2 = 2g(h_1 - h_2) v2=2g(h1−h2)v_2 = \sqrt{2g(h_1 - h_2)}
  4. Substitute the values and calculate.

    v2=2×9.8 m/s2×(20 m−5 m)v_2 = \sqrt{2 \times 9.8 \text{ m/s}^2 \times (20 \text{ m} - 5 \text{ m})} v2=2×9.8 m/s2×15 mv_2 = \sqrt{2 \times 9.8 \text{ m/s}^2 \times 15 \text{ m}} v2=294 m2/s2v_2 = \sqrt{294 \text{ m}^2/\text{s}^2} v2≈17.15 m/sv_2 \approx 17.15 \text{ m/s}

    The velocity of the body at height 5 m5 \text{ m} above the ground is 17.15 m/s\boxed{\text{17.15 m/s}}.

    2(c)(i) State three benefits that humans derive from the nitrogen cycle.

    • It provides nitrogen for the synthesis of proteins and nucleic acids (DNA, RNA), which are essential for all life.
    • It enriches soil fertility through the formation of nitrates, supporting agricultural productivity.
    • It enables the production of industrial chemicals like ammonia, used in fertilizers and explosives.

    2(c)(ii) State two human activities that affect the nitrogen cycle.

    • The extensive use of synthetic nitrogen fertilizers in agriculture.
    • The burning of fossil fuels, which releases nitrogen oxides into the atmosphere.

    2(d)(i) A colourless liquid in the laboratory was suspected to be water. Describe briefly a test to confirm that the liquid is water.
    Add a few drops of the liquid to anhydrous copper(II) sulfate. If the white powder turns blue, the liquid is confirmed to be water.

    2(d)(ii) State two stages in the large-scale water treatment.

    • Filtration: Water passes through layers of sand and gravel to remove smaller suspended particles.
    • Disinfection: Chemicals like chlorine are added to kill harmful microorganisms.

    3(a) The nuclides of sodium and fluorine are represented as 1123Na^{23}_{11}\text{Na} and 919F^{19}_{9}\text{F} respectively.
    (I) Protons in each atom.

    • For 1123Na^{23}_{11}\text{Na}: The atomic number (subscript) is 11, so it has 11 protons.
    • For $^{19
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