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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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2 years ago
A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
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Answer:

a) x = 0.098

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Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

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K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

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k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

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