The step by step experiment to determine the cubic expansivity of a liquid

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7 0

In accordance with the definition of density as r = m/V, in order to determine the density of matter, the mass and the volume of the sample must be known. The determination of mass can be performed directly using a weighing instrument. The determination of volume generally cannot be performed directly. Exceptions to this rule include · cases where the accuracy is not required to be very high, and · measurements performed on geometric bodies, such as cubes, cuboids or cylinders, the volume of which can easily be determined from dimensions such as length, height and diameter. · The volume of a liquid can be measured in a graduated cylinder or in a pipette; the volume of solids can be determined by immersing the sample in a cylinder filled with water and then measuring the rise in the water level. Because of the difficulty of determining volume with precision, especially when the sample has a highly irregular shape, a "detour" is often taken when determining the density, by making use of the Archimedean Principle, which describes the relation between forces (or masses), volumes and densities of solid samples immersed in liquid: From everyday experience, everyone is familiar with the effect that an object or body appears to be lighter than in air – just like your own body in a swimming pool. Figure 3: The force exerted by a body on a spring scale in air (left) and in water (right)

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Strike441 [17]

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Lady_Fox [76]

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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$