A child playing in a swimming pool realizes that it is easy to push a small, inflated ball under the surface of the water, whereas a large ball requires a lot of force. The child happens to have a styrofoam ball (the shape of which will not distort when it is submerged) that he forces under the surface of the water. If the child needs to supply 608 N to totally submerge the ball, calculate the diameter of the ball. The density of water is? w=1.000 g/cm3, the density of styrofoam is? foam=0.0950 g/cm3, and the acceleration due to gravity is?=9.81 m/s2.

Question

Physics

Tiara Stokes

4 March, 21:35

A child playing in a swimming pool realizes that it is easy to push a small, inflated ball under the surface of the water, whereas a large ball requires a lot of force. The child happens to have a styrofoam ball (the shape of which will not distort when it is submerged) that he forces under the surface of the water. If the child needs to supply 608 N to totally submerge the ball, calculate the diameter of the ball. The density of water is? w=1.000 g/cm3, the density of styrofoam is? foam=0.0950 g/cm3, and the acceleration due to gravity is?=9.81 m/s2.

+2

Answers (1)

Know the Answer?

Makai Reynolds · Answer 1

R = 25 cm

Explanation:

We can solve this problem using the Archimedes principle that the magnitude of hydrostatic thrust equal to the weight of the liquid dislodged

Let's start by reducing all units is to the SI system

ρ liq = 1000 kg / m3 (water)

ρ body = 0.0950 g / cm³ (1 kg / 1000g) 10⁶ cm³ / 1m³) = 95 kg/m³

We use Newton's second law for this equilibrium case

B - F - W = 0

B = ρ liq g V

ρ = m / V

m = ρ V

W = ρbody g V

ρliq g V body - F - ρbody g Vbody =

Vbody g (ρliq - ρbody) = F

V body = F / g (ρliq - ρbody)

V body = 608 / [9.81 (1000 - 95)

V body = 0.06855 m3

The volume of a sphere is

V = 4/3 pi R3

R = ∛ ¾ V / π

R = ∛ ¾ 0.06855 / π

R = 0.254 m

R = 25 cm