Respuesta :
Answer:
The diameter as calculated is 0.476 m
Solution:
As per the question:
Tension, T = [tex]5\times 10^{2} = 500\ N[/tex]
Density of water, [tex]1.00\times 10^{3}\ kg/m^{3}[/tex]
Density of the styrofoam, [tex]\rho_{foam} = 95.0\ kg/m^{3}[/tex]
Now,
To calculate the diameter, d of the ball:
With the help of the Archimedes's principle:
At equilibrium, upthrust equals the weight of the displaced liquid.
where
Mass, m = [tex]V\rho [/tex] (1)
where
V = Volume of ball
[tex]\rho = density[/tex]
Force, F = mg
From eqn (1):
Force , F = [tex]V\rho_{w} g[/tex] (1)
Now,
Net downward force is given by:
F' = F + T = [tex]V\rho_{foam} g[/tex] + T (2)
Thus equating (1) and (2):
F = F'
[tex]V\rho_{w}g = V\rho_{foam} g + T[/tex]
[tex]V = {T}{(\rho_{w} - \rho_{foam})}[/tex]
[tex]V = {500}{9.8(1.00\times 10^{3} - 95.0)} = 0.056\ m^{3}[/tex]
Also, we know that for ball:
[tex]V = \frac{4}{3}\pi R^{3}[/tex]
where
R = Radius of the ball
[tex]0.056 = \frac{4}{3}\pi R^{3}[/tex]
R = 0.238 m
Since, radius is known, the diameter is twice the radius:
Diameter, d = 2R = [tex]2\times 0.238 = 0.476\ m[/tex]
