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A metal block has a mass of 4 kg and a volume of 0.002 m². Find its density. 2. A force of 500 N acts on an area of 0.25 m². Calculate the pressure exerted. 3. Calculate the pressure at a depth of 8 m in water. (Take p = 1000 kg/m³, g = 9.8 m/s²

Physics

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Here are the solutions to the exercises:

1. Find its density.

  1. Identify the given values.

    Mass, m=4 kgm = 4 \text{ kg}
    Volume, V=0.002 m3V = 0.002 \text{ m}^3

  2. State the formula for density.

    Density (ρ\rho) is defined as mass per unit volume:

    ρ=mV\rho = \frac{m}{V}
  3. Substitute the values and calculate the density.

    ρ=4 kg0.002 m3\rho = \frac{4 \text{ kg}}{0.002 \text{ m}^3} ρ=2000 kg/m3\rho = 2000 \text{ kg/m}^3

    The density of the metal block is:

    2000 kg/m3\boxed{\text{2000 kg/m}^3}

    2. Calculate the pressure exerted.

  4. Identify the given values.

    Force, F=500 NF = 500 \text{ N}
    Area, A=0.25 m2A = 0.25 \text{ m}^2

  5. State the formula for pressure.

    Pressure (PP) is defined as force per unit area:

    P=FAP = \frac{F}{A}
  6. Substitute the values and calculate the pressure.

    P=500 N0.25 m2P = \frac{500 \text{ N}}{0.25 \text{ m}^2} P=2000 N/m2P = 2000 \text{ N/m}^2 P=2000 PaP = 2000 \text{ Pa}

    The pressure exerted is:

    2000 Pa\boxed{\text{2000 Pa}}

    3. Calculate the pressure at a depth of 8 m in water.

  7. Identify the given values.

    Depth, h=8 mh = 8 \text{ m}
    Density of water, ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3
    Acceleration due to gravity, g=9.8 m/s2g = 9.8 \text{ m/s}^2

  8. State the formula for pressure in a fluid at a certain depth.

    The pressure (PP) at a depth hh in a fluid is given by:

    P=ρghP = \rho g h
  9. Substitute the values and calculate the pressure.

    P=(1000 kg/m3)×(9.8 m/s2)×(8 m)P = (1000 \text{ kg/m}^3) \times (9.8 \text{ m/s}^2) \times (8 \text{ m}) P=9800×8 PaP = 9800 \times 8 \text{ Pa} P=78400 PaP = 78400 \text{ Pa}

    The pressure at a depth of 8 m in water is:

    78400 Pa\boxed{\text{78400 Pa}}
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