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19 April, 01:06

The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.200 mm in radius, what pressure is exerted on the record in kPa

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  1. 19 April, 04:23
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    Pressure exerted by the needle = 78.037 kPa

    Explanation:

    Pressure is defined as the force exerted on a surface per unit area. The SI unit of pressure is Pa (N. m⁻²).

    The force exerted by the phonograph needle is equal to the gravitational force.

    Thus,

    F=m*g

    Where,

    F is the force

    m is the mass of the body

    g is the acceleration due to gravity

    Also, g = 9.81 ms⁻²

    Given, mass of the needle = 1.00 g

    The conversion of g into kg is shown below:

    1 g = 10⁻³ kg

    Thus, mass of the needle = 1*10⁻³ kg

    Thus, F = 1*10⁻³*9.81 N = 9.81*10⁻³ N

    Ares of the circle = πr²

    Given : Radius of tip of the needle = 0.200 mm

    The conversion of mm into m is shown below:

    1 mm = 10⁻³ m

    Thus, mass of the needle = 0.2*10⁻³ m

    Thus, Area of the tip of the needle = π * (0.2*10⁻³) ² m² = 1.2571*10⁻⁷ m²

    So, Pressure = F/A

    P = (9.81*10⁻³) / 1.2571*10⁻⁷ Pa = 7.8037*10⁴ Pa

    The conversion of Pa into kPa is shown below:

    1 Pa = 10⁻³ kPa

    Thus, pressure exerted by the needle = 7.8037*10⁴*10⁻³ kPa = 78.037 kPa
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