A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera strobe light.

Question

A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera strobe light. What radiation pressure is produced on the cardboard by the light?

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Bình An 2 months 2021-08-02T09:39:29+00:00 2 Answers 1 views 0

Answers ( )

    0
    2021-08-02T09:40:31+00:00

    Answer:

    2.7×10⁻⁸ N/m²

    Explanation:

    Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:

     p_{rad} = \frac{I}{c}

    Where:

     p_{rad}: is the radiation pressure

    I: is the intensity of the light = 8.1 W/m²

    c: is the speed of light = 3.00×10⁸ m/s

    Hence, the radiation pressure is:

     p_{rad} = \frac{I}{c} = \frac{8.1 W/m^{2}}{3.00 \cdot 10^{8} m/s} = 2.7 \cdot 10^{-8} N/m^{2}

    Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7×10⁻⁸ N/m².

    I hope it helps you!

    0
    2021-08-02T09:41:24+00:00

    Answer:

    The answer is 2.7×10⁻⁸ Pa

    Explanation:

    The expression for total absorption is:

    Pr = I/c

    Where I = intensity of light

    c = velocity of light

    Replacing:

    Pr = 8.1/3×10⁸ = 2.7×10⁻⁸ Pa

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