Respuesta :
Solution :
Mass of the sphere, m = 10 kg
Diameter, D = 300 mm = 0.3 m
Volume of the sphere is [tex]$V=\frac{4}{3} \pi \left(\frac{0.3}{2}\right)^3$[/tex]
[tex]$V=0.01414 \ m^3$[/tex]
So the volume of displaced water, [tex]$V_w=V = 0.01414 \ m^3$[/tex]
Additional mass, [tex]$m_w = \frac{1}{2} \ \rho_w\times v_w$[/tex]
[tex]$=\frac{1}{2} \times 1000 \times 0.01414$[/tex]
[tex]$=7.0686 \ kg$[/tex]
So the total mass, [tex]$M = m_w+m$[/tex]
= 7.0686 + 10
= 17.0686
Force required, F = Ma
[tex]$F=17.0686 \times 10$[/tex]
= 170.686 N
The force that is necessary to accelerate when the acceleration rate of 10 m/s2 in the horizontal direction should be 170.686 N.
How to calculate the force?
Since Mass of the sphere, m = 10 kg
Diameter, D = 300 mm = 0.3 m
Now volume of the sphere is
V = 4/π (diameter/3)^3
V = 4/3π(0.3/2)^3
= 0.01414m^3
Now the additional mass is
= 1/2* 1000 * volume
= 1/2 *1000*0.01414
= 7.0686
So, the total mass is
= Additional mass + force
= 7.0686 + 10
= 17.0686
Now the force is
= 17.0686 * 10
= 170.686
Hence, The force that is necessary to accelerate when the acceleration rate of 10 m/s2 in the horizontal direction should be 170.686 N.
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