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Dafna11 [192]
4 years ago
11

An ice cube measures 1.38 in. on each edge and weighs 1.39 oz. What is the density of the ice cube in g/cm^3

Physics
1 answer:
brilliants [131]4 years ago
4 0

Answer:

0.91 g/cm³

Explanation:

Density: This can be defined as the ratio of mass of an object to its volume.

The mathematical expression for density is given as,

D = m/v .............................. Equation 1

Where D = Density of the ice cube, m = mass of the ice cube, v = volume of the ice cube.

v = l³

Where l = length of each side of the cube.

l = 1.38 in = (1.38×2.54) cm = 3.51 cm

v = 3.51³

v = 43.24 cm³

Given: m = 1.39 oz = (1.39×28.35) g = 39.41 g

Substitute into equation 1

D = 39.41/43.24

D = 0.91 g/cm³

Hence the density of the ice cube = 0.91 g/cm³

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The rotational inertia about the end of a uniform rod=\frac{1}{3}ML^2

Kinetic energy at the level of center of mass of rod below the pivot=\frac{1}{2}I\omega^2

Kinetic energy at the level of center of mass of rod above the pivot=0

Potential energy at the level of center of mass of rod above the pivot=mgh

We have to find the center of mass ( in terms of g and L).

According to conservation of law of energy

Initial P.E+Initial K.E=Final P.E+Final K.E

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Where K.E=\frac{1}{2} I\omega^2

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Substitute the values then we get

MgL=\frac{1}{2}\times \frac{1}{3}ML^2\omega^2

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Now, we know that \omega=\frac{v}{r}, r=\frac{L}{2}

Substitute the values then we get

\frac{v^2}{(\frac{L}{2})^2}=\frac{6g}{L}

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Hence, the speed of its center of mass =\sqrt{\frac{3}{2}gL}

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