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A 1.50 kg rock whose density is 4700 kg/m3 is suspended by a string such that half of the rock's volume is under water.
What is the tension in the string?(In N)

Respuesta :

Answer:

Tension T = 13.14N

Explanation:

Given:

Mass of rock m = 1.50kg

Density of rock p = 4700kg/m^3

Volume of rock V = mass/density = m/p

V = 1.50kg/4700kg/m3 = 3.19×10^-4m3

Taking the summation of forces acting on the rock;

T-W+Fb = 0

T = W - Fb .....1

T = tension

W = weight of rock

Fb = buoyant force

Fb = pw(0.5V)g = density of water × Volume under water×™ acceleration due to gravity

g = 9.8m/s^2

T = mg - pw(0.5V)g

T = 1.50×9.8 - 1000kg/m^3 ×0.5(3.19 × 10^-4) × 9.8

T = 13.14N

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