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Maslowich
3 years ago
8

A penny has a mass of 2.50 g, a diameter of 19.55 mm, and a thickness of 1.55 mm. Calculate the density of the material from whi

ch the penny is made.
Physics
1 answer:
Dmitry [639]3 years ago
4 0
Density = (mass) divided by (volume)

We know the mass (2.5 g).  We need to find the volume.

The penny is a very short cylinder.
The volume of a cylinder is (π · radius² · height).
The penny's radius is 1/2 of its diameter = 9.775 mm.
The 'height' of the cylinder is the penny's thickness = 1.55 mm.

Volume  =  (π) (9.775 mm)² (1.55 mm)

             =  (π) (95.55 mm²) (1.55 mm)

             =  (π) (148.1 mm³)

             =        465.3 mm³

We know the volume now.  So we could state the density of the penny,
but nobody will understand what we have.  Here it is:

         mass/volume = 2.5 g / 465.3 mm³  =  0.0054 g/mm³  .

Nobody every talks about density in units of ' gram/(millimeter)³ ' .
It's always ' gram / (centimeter)³ '.
So we have to convert our number for the volume.

                         (0.0054  g/mm³)  x  (10 mm / cm)³

                 =      (0.0054 x 1,000)  g/cm³

                 =          5.37  g/cm³  .

This isn't actually very close to what the US mint says for the density
of a penny, but it's in a much better ball park than 0.0054 was.
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A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
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Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

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Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

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The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

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             = 0.1236 m

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      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

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          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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