Please who can give me the definition of density in terms of Archimedes principle.

Physics

2 answers:

VladimirAG [237]2 years ago

8 0

$\large\bold{\underline{\underline{Answer:-}}}$

The weight of the fluid displaced is equal to the buoyant force on a submerged object. The mass divided by the volume thus determined gives a measure of the average density of the object.

serg [7]2 years ago

8 0

The density, of a substance is its mass per unit volume. The symbol most often used for density is ρ, although the Latin letter D can also be used. Mathematically, density is defined as mass divided by volume: {\displaystyle \rho ={\frac {m}{V}}} where ρ is the density, m is the mass, and V is the volume.

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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

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Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$