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Tcecarenko [31]
2 years ago
12

One beaker contains 100 mL of pure water and second beaker contains 100 mL of seawater. The two beakers are left side by side on

a lab bench for one week. At the end of the week, the liquid in both beakers has decreased, however, the level has decreased more in one of the beakers than in the other. Explain why one of them has decreased more by filling in the following sentence:
Physics
2 answers:
Harman [31]2 years ago
7 0

Answer: The beaker containing pure water has decreased more.

Explanation:

In both cases, the decrease of water level is due to evaporation. We know that evaporation is a surface phenomenon. In the case of salt water, the salt molecules somewhat hinders the evaporation process of the water molecules and hence the salt water evaporates at a slower rate than pure water.

Hence, pure water level falls more.

marishachu [46]2 years ago
4 0

Answer:

The answer is:

The presence of the dissolved salt in the sea water means that there is a lower number of water molecules at the surface. This therefore means that the sea water will have lower rate of vaporization than the pure water which is why the water level in the sea water beaker is higher by the end of the weak.

Explanation:

Pure water is characterized by being only made up of water molecules. Only those molecules will exist on its surface. On the other hand, seawater is made up of different types of ions, other chemical compounds, etc. In the glass that contains sea water on its surface there will be different dissolved molecules, in addition to the water molecules. In this way, the number of water molecules will be less compared to pure water. Regarding the evaporation rate, in pure water it is higher than that of sea water. This is why the glass that only contains pure water will evaporate much faster than the glass that contains sea water.

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mario62 [17]

Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

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For the electric field to be zero at the location

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\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}

\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}

\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}

x = - 1.4 m

6 0
2 years ago
Use the drop-down menus to complete the statements. The atoms in a solid . The arrangement of atoms in a solid causes it to have
Natalija [7]

Answer:The atoms in a solid  .

remain in fixed position

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Explanation:

8 0
2 years ago
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2 years ago
How do air mass conditions ahead of the squall line support the development of new cell?
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7 0
2 years ago
A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s
Anastaziya [24]

Answer:

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as part of rest v₀ = 0

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