A 4 cm diameter “bobber” with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber, and from t

Question

A 4 cm diameter “bobber” with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber, and from the bottom hangs a 10 gram lead weight. The density of lead is 11.3 g/cm3. What fraction of the bobber’s volume is submerged, as a percent of the total volume

in progress 0
Gerda 4 years 2021-08-26T20:52:24+00:00 1 Answers 92 views 0

Answers ( )

  1. Adela
    0
    2021-08-26T20:53:36+00:00
    Reply

    Answer:

    Explanation:

    Calculate the volume of the lead

    V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}

    Now calculate the bouyant force acting on the lead

    F_L = Vpg

    F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N

    This force will act in upward direction

    Gravitational force on the lead due to its mass  will act in downward direction

    Hence the difference of this two force

    T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N

    If V is the volume submerged in the water then bouyant force on the bobber is

    F_B=V'pg

    Equate bouyant force with the tension and gravitational force

    F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}

    Now Total volume of bobble is

    \frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\

    =\large\boxed{4.52 \%}

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )