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  2. 3 days ago · A sphere is a perfectly round geometrical 3-dimensional object. It can be characterized as the set of all points located distance r r (radius) away from a given point (center). It is perfectly symmetrical, and has no edges or vertices. A sphere with radius r r has a volume of \frac {4} {3} \pi r^3 34πr3 and a surface area of 4 \pi r^2 4πr2.

  3. 1 day ago · Archimedes' Principle formula: Archimedes’ Principle is an important topic taught to students from the 9th standard. It is a part of the Physics syllabus. ... Given: A metal sphere with a volume ...

  4. 1 day ago · The study of the relationships between the sides and angles of triangles drawn on a sphere’s surface is known as spherical trigonometry. By using trigonometric concepts in non-planar geometry, it deals with the measurement and computation of angles, distances, and areas on spherical surfaces. Spherical Trigonometry.

  5. 3 days ago · Formula. The formula to calculate the density of a sphere involves the volume of a sphere and its mass: \[ D = \frac{m}{\frac{4}{3}\pi r^3} \] where: \(D\) is the density in kilograms per cubic meter (kg/m³), \(m\) is the mass of the sphere in kilograms, \(r\) is the radius of the sphere in meters. Example Calculation

  6. 3 days ago · To determine the volume of a sphere, we use the formula 4/3πr^3. So, to find the volume of a hollow sphere, we must subtract the volume of the hollow region from the volume of the overall sphere. Let's call the radius of the overall sphere r1 and the radius of the hollowed region r2.

  7. 2 days ago · Determine the formula for the volume of a sphere. The volume \( V \) of a sphere is given by: \[ V = \frac{4}{3} \pi r^3 \] where \( r \) is the radius of the sphere. Step 2/11 Given the volume of the sphere is 7253 cm\(^3\), set up the equation: \[ 7253 = \frac{4}{3} \pi r^3 \] Step 3/11

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