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)

Question

Physics

Karly Chan

13 October, 22:01

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)

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Answers (1)

Know the Answer?

Octavio Coffey · Answer 1

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

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)

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)