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A cork is held at the bottom of a bucket of water by a piece of string. The actual depth of the cork is 0.6 m below the surface of the water. The density of water is ρ = 1000 kg/m3 and the acceleration of gravity is g = 9.8 m/s2 . 0.6 m If the density of the cork is 150 kg/m3 and the volume of the cork is 2 cm3, then what is the tension in the string?

Respuesta :

Answer:

tension on the string will be 0.0167 N

Explanation:

ρ = 1000 kg/m³

density of the cork = 150  kg/m³

volume of the cork = 2 cm³

tension on cork (T)  will be acting in down ward direction.

buoyant force (Fₙ) will act in upward direction.

and weight (W) will act in downward direction.

T = Fₙ - W

[tex]T= \rho Vg- mg\\T=\rho V_{cork} g - \rho_{cork} V_{cork}g\\T=1000\times 2\times 10^{-6}\times 9.81 - 150\times 2\times 10^{-6}\times 9.81\\ T= 0.0167 N[/tex]

hence tension on the string will be 0.0167 N

onyebuchinnaji

The tension on the string in the bucket of water is 0.017 N .

The given parameters;

  • depth of the cork, h = 0.6 m
  • density of the water, ρ = 1000 kg/m³
  • density of the cork, = 150 kg/m³
  • volume of the cork, V = 2 cm³

The tension on the string is the buoyant force experienced by the cork is calculated as follows;

[tex]T =\rho _{w} Vg- \rho _{cork} Vg\\\\T = 1000 \times (2\times 10^{-6}) \times 9.8 \ \ - \ \ 150 \times (2\times 10^{-6}) \times 9.8\\\\T = 0.017 \ N[/tex]

Thus, the tension on the string is 0.017 N.

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