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When a body is accelerated under water, some of the surrounding water is also accelerated. This makes the body appear to have a larger mass than it actually has. For a sphere at rest this added mass is equal to the mass of one half of the displaced water. Calculate the force necessary to accelerate a 10 kg, 300-mm-diameter sphere which is at rest under water at the acceleration rate of 10 m/s2 in the horizontal direction. Use H2O

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.

Learn more about acceleration here: https://brainly.com/question/18555689

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