Samples Physics Archimedes’ Principle Lab

Archimedes’ Principle Lab

540 words 2 page(s)

1) A solid sphere has a radius of 5 cm and a mass of 350 grams. What is its average density? A hollow sphere has the same radius and mass. What is its average density?
a) Average density solid sphere= 〖Mass〗_TOTAL/〖Volume〗_TOTAL = (350 grams)/((4/3)*π*5^3 〖cm〗^3 )= 0.668 grams/〖cm〗^3
b) Average density hollow sphere= 〖Mass〗_TOTAL/〖Volume〗_TOTAL = (350 grams)/((4/3)*π*(5^3-〖r_in〗^3 ) 〖cm〗^3 )
I assumed the outer radius and inner radius of the hollow sphere as 5 cm (given) and rin cm respectively. Irrespective of the inner radius value, the density of the hollow sphere will always be greater than its solid counterpart.

2) Write an expression for the weight of an object in terms of its density.
Terminology: Mass is m, Weight W, density d, Volume V and acceleration due to gravity is g.
W = m*g (unit: (kg*m)/s^2 )
d=m/V (unit: kg/m^3 )
So, m= W/g
Therefore, W= d*V*g

Need A Unique Essay on "Archimedes’ Principle Lab"? Use Promo "custom20" And Get 20% Off!

Order Now

3) Explain why the buoyant force is greater if you are swimming in the ocean versus fresh water.
Sea water has a density of about 1025 kg/m^3 >Fresh water has a density of about 1000 kg/m^3
The greater density of ocean compared to that of fresh water causes a greater buoyant force application on the body of the swimmer. The greater density of the ocean water is due to presence of dissolved salts that adds to the existing upward force by water molecules.

4) A straight sided bucket is half full of water. The bucket is round with radius of 20cm, and it is 50cm tall. How much does the water level go up if you drop a steel sphere of radius 5cm into the bucket? How much does the water level go up if you drop in a 5cm radius sphere made out of Idaho white pine?
a) Steel has a density of 8,050 kg/m3 so it will be completely immersed in water. Therefore, water level rise (volume rise in a half-filled bucket) is equal to the volume of the steel sphere.
Volume of water in the bucket= π*r^2*h= π*〖20〗^2*(h) 〖cm〗^3;where h is water level
Volume of steel sphere=Volume of water in the bucket= (4/3)*π*5^3=523.598 〖cm〗^3
Therefore, h= 0.416 cm is the water level rise in case of steel
a) Idaho white pine has a density of 530 kg/m3 so it displaces water equal to its weight.
Volume of water in the bucket= π*r^2*h= π*〖20〗^2*(h) 〖cm〗^3;where h is water level
Weight of Idaho pine=d*V=530* (4/3)*π*(5/100)^3=0.277 kg
277 grams of wood is equivalent to 277 grams of water which is also 277 cm3 of water
Volume of water in the bucket= π*r^2*h= π*〖20〗^2*(h) 〖cm〗^3=277 cm3
Therefore, h=0.220 cm is the water level rise in case of pine.

5) You are swimming in the indoor pool at the Holiday Inn Moon-Base in the year 2041. Will you float higher, lower or the same in the pool compared to swimming on Earth? Explain.
It will be the same as that of Earth because the relative density between the body of the swimmer and water in the pool will remain same as that of Earth. The buoyant force applied by water would be much smaller in moon’s gravity but the weightless observed by the swimmer will compensate for the lower buoyant force application.