How Much Cork Is Needed to Keep a Child Afloat on a Water Slide?

[Dragoon]

this is what the problem states i need help getting going
An ameusement park is planning to build a water slid, and the park owners would like to use giants corcks (pc=.28 g/cm3) as floats for young children. what volumer of cork is needed to keep 20 percent by volume of 40 kg child above the water at the base of the water slide (assuming the density of a child is 1.01 g/cm3.

 

[marlon]

well what do you have up till now. Maybe some hints : there are a few ways to solve this one. Think of stuff like pressure and the relation between pressure-difference and height difference : p = p' + ro*g*(h'-h)... where ro is the density of the surrounding matter. Also consider Archimedes' law...

Please do not cross-post...

regards
marlon

 

[wais]

Hello,

Thank you for reaching out for help with your fluid problem. It seems like you are trying to calculate the volume of cork needed to keep a 40 kg child afloat on a water slide. To solve this problem, we will need to use the formula for buoyancy, which states that the buoyant force (upward force) on an object is equal to the weight of the fluid that it displaces.

First, let's convert the density of the cork from g/cm3 to kg/m3, since the density of the child is also given in kg/m3. This can be done by multiplying the density by 1000. So, the density of cork is 280 kg/m3.

Next, we can calculate the volume of cork needed to displace 20% of the volume of the child. This can be done by multiplying the volume of the child (40 kg / 1.01 g/cm3 = 39.6 L) by 0.20. This gives us a volume of 7.92 L.

Now, we can use the density of cork and the volume we calculated to find the mass of cork needed. The mass of cork can be found by multiplying the volume (converted to m3) by the density. So, the mass of cork needed is 0.00792 m3 x 280 kg/m3 = 2.22 kg.

Therefore, the volume of cork needed to keep 20% of the volume of a 40 kg child above water at the base of the water slide is 7.92 L, or approximately 2.22 kg. I hope this helps you get started on solving your problem. Let me know if you need further assistance. Good luck!