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Sever21 [200]
3 years ago
7

You may have been surprised to learn that Olympic gold medals are not made from solid gold, but instead have a coating of • Save

d gold on the outside.
To see a possible reason why, determine the value of the medal the size (not mass) of the Olympic gold medal if it were made of solid gold. Hint: As of mid-2018, the cost of gold is about $40 per gram.
Physics
1 answer:
Dimas [21]3 years ago
5 0

Answer:

A gold medal has the (minimum) dimensions of:

diameter = 60mm

thickness = 3mm

So we will work with those dimensions.

The medal is then a cyinder of diameter

D = 60mm = 6cm

and height:

H = 3mm = 0.3cm

Remember that the volume of a cylinder is:

V = pi*(D/2)^2*H

where pi = 3.14

Then the volume of a medal is:

V = 3.14*(6cm/3)^2*0.3cm = 3.768 cm^3

The density of the gold in g/cm^3 is:

d = 19.3 g/cm^3

And remember that:

density = mass/volume

So, if the volume is 3.768 cm^3

Then the mass will be:

mass = density*volume =  19.3 g/cm^3*3.768 cm^3 = 72.7 g

So, a single gold medal would weight 72.7 grams

And each gram of gold costs $40

Then the total cost of the gold medal would be:

value = $40*72.7 = $2,908

Now, if yo think that in the Olympics there are 35 sports (a lot with a large number of players) and near 50 disciplines, they need a lot of gold medals.

And each gold medal costs $2,908

So the total cost (only for the gold medals, ignoring the others) would be to high.

This is why the gold medals are made mostly of silver.

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a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
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D/H =15

Explanation:

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       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

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       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
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        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

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The number of ice blocks that can be stored in the freezer is 80 blocks of ice.

<h3>Average thickness of a sheet of the paper</h3>

The average thickness of a sheet of the paper is calculated as follows;

average thickness = 6 mm/60 sheets = 0.1 mm /sheet

Thus, the average thickness of a sheet of the paper is 0.1 mm.

<h3>Volume of each block of ice</h3>

Volume = 10 cm x 10 cm x 4 cm

Volume = 400 cm³

<h3>Volume of the freezer</h3>

Volume = 40 cm x 40 cm x 20 cm = 32,000 cm³

<h3>Number of ice blocks that can be stored</h3>

n = 32,000 cm³/400 cm³

n = 80 blocks of ice

Thus, the number of ice blocks that can be stored in the freezer is 80 blocks of ice.

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