centroid
center of gravity
center of gravity
the centroid the point at which one can balance the triangle
2 apex fill-in-the-blank Q's: 1) The center of gravity of a trianglular solid with uniform thickness and density is at the intersection of the medians of the triangle. 2) The center of gravity of a triangular solid with uniform thickness and density is the centroid. medians the point at which one can balance the triangle. and the point shared by a triangle's medians. medians centroid Medians i got centriod?
the point shared by a triangle's medians the point at which one can balance the triangle
center of gravity
center of gravity
the centroid the point at which one can balance the triangle
2 apex fill-in-the-blank Q's: 1) The center of gravity of a trianglular solid with uniform thickness and density is at the intersection of the medians of the triangle. 2) The center of gravity of a triangular solid with uniform thickness and density is the centroid. medians the point at which one can balance the triangle. and the point shared by a triangle's medians. medians centroid Medians i got centriod?
Archimedes showed that the point where the medians are concurrent is the center of gravity of a triangular shape of uniform thickness and density.
If the cube is uniform ( ie it has uniform density) then the geometric center of the cube is its center of gravity.
the point shared by a triangle's medians the point at which one can balance the triangle
The center of gravity of a triangle can be found by adjusting the thickness. You also need to find the density at the intersection.
No, the center of gravity of a meterstick is not always located at the 50-cm mark. The center of gravity of an object is the point where its weight is considered to act. For a uniform meterstick, the center of gravity will indeed be at the 50-cm mark because of its uniform density distribution, but if the density distribution is not uniform, the center of gravity could be located at a different point.
the area of the triangular face (0.5 x base x height) times the length of the prism * * * * * No. That will only give the volume which is not the same as the mass. You will either need to assume that the prism is of uniform density. In that case, you multiply its volume by the density. Alternatively, you follow Archimedes' principle to determine the density or even the weight of the prism. Then you need to convert to mass by dividing by the force of gravity. Not as easy as the first answer wrongly made it look.
Assuming the ball is a perfect sphere of uniform density, and is suspended from a massless string, the centre of gravity is in the centre of the ball.
The center of gravity is the theoretical point where all the body weight is concentrated or the theoretical point about which the body weight is evenly distributed. If a body is of uniform density and has a symmetrical shape the center of gravity is in the geometric center. If the object is not symmetrical and does not have uniform density, it is more difficult to describe the location of its center of gravity.