Volume of sphere = volume of rise in water.
volume of sphere = 4/3 x π x radius3
volume of rise in water = π x (diameter/2)2 x rise
→ 4/3 x π x radius3 = π x (diameter/2)2 x rise
→ radius = 3√(3/4 x (diameter/2)2 x rise)
= 3√(3/4 x (10 cm/2)2 x 3 cm)
= 3√(3/4 x (10 cm/2)2 x 3 cm)
≈ 3.83 cm
To calculate the rise in the water level when a spherical ball is immersed in water in a vertical cylinder, you can use the Archimedes' principle, which states that the buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object. The volume of water displaced by the ball is equal to the volume of the ball. Calculate the volume of the ball using the formula for the volume of a sphere, then use this volume to determine the rise in water level.
Diameter is the width of a circular or spherical figure or object. Something 180 inches in diameter is also 180 inches in width or thickness.
It is not the weight of the immersed object but the volume of the object would affect the buoyant force on the immersed object because the buoyant force is nothing but the weight of the displaced liquid whose volume is equal to that of the immersed object.
-- volume of the object immersed in fluid -- density of the fluid in which the object is immersed
measure the circumference, c. Since c = pi x d where d = diameter pi = 3.1415926535897932384676..... diameter = circumference / pi
None. I am sure tat if you measured it to a very high degree of accuracy it would not be exactly 120 km. Furthermore, there is not an object that is perfectly spherical so the diameter will vary from point to point.
The diameter of the internal measurement of the cylinder.A mathematical cylinder only has one diameter.A real world object that is a cylinder (e.g. a pipe) has a wall with a thickness and thus the internal and external diameters will differ by twice the thickness of the wall.
The most spherical object in the universe is believed to be the star Kepler 11145123, which is nearly perfect in its spherical shape.
The weight loss of an object when immersed in a liquid is due to the buoyant force acting on the object. This force is equal to the weight of the liquid displaced by the object. As a result, the apparent weight of the object is reduced when immersed in a liquid.
Volume displacement is the method used to find the volume of small or irregularly shaped objects by noting the difference in the level of liquid before and after after immersing an object into a graduated cylinder or beaker of liquid. For a small object immersed in a graduated cylinder or beaker, the volume displaced by the object can be read directly from the scale on the container.
A diameter is the distance from one side of a spherical or circular object, through the centre, to the other. It can be as small as an atomic sub-particle or as large as a red super-giant star. There is absolutely no way that 6 millimetres can be "converted" to diameter.
When an object is immersed in water, it means it is completely surrounded or covered by the water.