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3 years ago

Why does the mirror get misty when someone uses the shower ??

2 answers:

const2013 [10]3 years ago

6 0

The steam from the shower's hot water comes into contact with a cool surface(the mirror) and the steam condenses from a gas into a liquid on the mirror.

Colt1911 [192]3 years ago

3 0

The steam from the shower's hot water comes into contact with a cool surface(the mirror) and the steam condenses from a gas into a liquid on the mirror.

Explanation:

The fog within the mirror is that the condensation of water vapor because it touches a colder surface. By running cold water you only quiet down the bathtub tub and everything around it. currently the vapor coming back from the recent shower can principally condense right there and can not reach the mirror.

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Paha777 [63]

Answer:

A) $V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

B) $d = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

Explanation:

A) Using the Archimedes' force we can find the weight of water displaced:

$W_{d} = W_{a} - W_{w}$

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$W_{a}$ : is the weight of the block in the air = 20.1 N

$W_{w}$ : is the weight of the block in the water = 15.3 N

$W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N$

Now, the mass of the water displaced is:

$m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg$

The volume of the block can be found using the mass of water displaced and the density of the water:

$V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

B) The density of the block can be found as follows:

$d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

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