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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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Springfield's "classic rock" radio station broadcasts at a frequency of 102.1 mhz. what is the length of the radio wave in meter
Mila [183]
The frequency of the radio wave is:
f=102.1 MHz = 102.1 \cdot 10^6 Hz

The wavelength of an electromagnetic wave is related to its frequency by the relationship
\lambda= \frac{c}{f}
where c is the speed of light and f the frequency. Plugging numbers into the equation, we find
\lambda= \frac{3 \cdot 10^8 m/s}{102.1 \cdot 10^6 Hz}= 2.94 m
and this is the wavelength of the radio waves in the problem.
7 0
2 years ago
A ball of mass m and radius R is both sliding and spinning on a horizontal surface so that its rotational kinetic energy equals
spin [16.1K]

Answer:

\frac{v_{cm}}{\omega} = 1.122\cdot R

Explanation:

According to the statement of the problems, the following identity exists:

K_{t} = K_{r}

\frac{1}{2}\cdot m \cdot v_{cm}^{2} = 0.63\cdot m \cdot R^{2} \cdot \omega^{2}

After some algebraic handling, the ratio is obtained:

\frac{v_{cm}^{2}}{\omega^{2}}=1.26\cdot R^{2}

\frac{v_{cm}}{\omega} = 1.122\cdot R

4 0
2 years ago
First find ∮RB⃗ ⋅dl⃗ , the line integral of B⃗ around a loop of radius R located just outside the left capacitor plate. This can
Allisa [31]

Answer:

the expression of current in the loop enclosed to the left of the capacitor plate is

I(t) = \frac{1}{\mu_0}\int B. dL

Explanation:

As we know by Ampere's law that line integral of magnetic field around a closed loop is proportional to the current enclosed in the path

So we will have

\int B. dL = \mu_0 I(t)

so we have

I(t) = \frac{1}{\mu_0}\int B. dL

so above is the expression of current in the loop enclosed to the left of the capacitor plate

5 0
2 years ago
11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0
Greeley [361]

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or  it is filled with water , in both the cases pressure centre will lie at the centre of the car .

7 0
2 years ago
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

6 0
2 years ago
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