Answer:
0.26
Explanation:
Given that :
Diameter of ball = 3.81 cm = 3.81/100 = 0.0381 m
Radius (r) = 0.0381 / 2 = 0.01905 m
Average density of ball (Db) = 0.0842 g/cm³ = (0.0842 / 1000)kg / 10^-6 = 0.0842/ 1000 * 10^6 = 84.2kg/m³
Density of water (Dw) = 1000kg/m³
Volume of hollow ball: (4/3) * pi * r³
V = (4/3) * π * 0.01905^3
V = 0.0000289583 m³
Required force = (Dw * V * g) - (Db * V * g)
= (1000 * 0.0000289583 * 9.8) - (84.2 * 0.0000289583 * 9.8)
= 0.259896109172
= 0.2598 N
To determine the force that must be applied to the hollow ball to hold it stationary while it is completely submerged under water is 0.2243 Newton.
Given the following data:
Density of water = 1000 [tex]kg/m^3[/tex]
Conversion:
Diameter of hollow ball = 3.81 centimeters to meters = [tex]\frac{3.81}{100} = 0.0381\;m[/tex]
Density of hollow ball = 0.0842 [tex]g/cm^3[/tex] to [tex]kg/m^3[/tex] = [tex]\frac{0.0842 \times 10^6}{1000} = 84.2 \;kg/m^3[/tex]
To determine the force that must be applied to the hollow ball to hold it stationary while it is completely submerged under water:
First of all, we would calculate the volume of the hollow ball by using the formula:
[tex]Volume = \frac{4}{3} \pi r^3\\\\Volume = \frac{4}{3} \times 3.142 \times 0.01905^3\\\\Volume = 4.19 \times 0.0000069\\\\Volume = 0.000029 \;m^3[/tex]
Next, we would find the force required for water and the hollow ball:
For water:
[tex]F_w = D_wVg\\\\F_w = 1000 \times 0.000029 \times 9.8\\\\F_w = 0.2842\; Newton[/tex]
For hollow ball:
[tex]F_h = D_hVg\\\\F_h = 84.2 \times 0.000029 \times 9.8\\\\F_h = 0.0239\; Newton[/tex]
Now, we can find the force that must be applied to the hollow ball:
[tex]Force = F_w - F_h\\\\Force = 0.2482 - 0.0239[/tex]
Force = 0.2243 Newton
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