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hjlf
4 years ago
5

When solid surfaces slide over each other, the kind of friction that occurs is called sliding friction.

Physics
1 answer:
LekaFEV [45]4 years ago
6 0

Answer:

True

Explanation:

A sled sliding across snow or ice. Skis sliding against snow. A person sliding down a slide is an example of sliding friction. A coaster sliding against a table. Two cards in a deck sliding against each other. All of these are examples of sliding friction.

Hope this helps! :)

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4 years ago
The mass of a copper atom is 1.06 10-25 kg, and the density of copper is 8 920 kg/m3 . (a) determine the number of atoms in 1 cm
Andreyy89

Mass of copper atom is 1.06\times 10^{-25}kg and density of copper atom is 8920 kg/m^{3}.

First calculate the volume from density and mass as follows:

V=\frac{m}{d}=\frac{1.06\times 10^{-25} kg}{8920 kg/m^{3}}=1.188\times 10^{-29}m^{3}

Now, convert m^{3} to cm^{3},

1 m^{3} = 10^{6} cm^{3}

Thus,

1.188\times 10^{-29}m^{3}=1.188\times 10^{-23}cm^{3}

Now,

1.188\times 10^{-23}cm^{3}\rightarrow 1.06\times 10^{-25}kg

Thus,

1cm^{3}\rightarrow \frac{1.06\times 10^{-25}kg}{1.188\times 10^{-23}kg}=0.00892 kg

or,

m=8.92 g

The molar mass of copper is 63.54 g/mol thus, number of moles can be calculated as:

n=\frac{m}{M}=\frac{8.92 g}{63.54 g/mol}=0.1403 mol

Now, in 1 mol of Cu there are 6.023\times 10^{23} atoms, number of atoms in 0.1403 mol will be:

N=0.1403\times 6.023\times 10^{23}=8.45\times 10^{22} atoms

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