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LuckyWell [14K]
1 year ago
5

When collecting a gas over water, the gas is always a mixture of the gas collected and water vapor.

Physics
1 answer:
bagirrra123 [75]1 year ago
7 0

When collecting a gas over water, the gas is always a mixture of the gas collected and water vapor. This statement is True

Water vapour or aqueous vapour is the name for the gaseous state of water. It is a particular kind of water condition inside the hydrosphere. Both the boiling and evaporation of liquid water as well as the sublimation of ice can produce water vapour. Water vapour is transparent, like the bulk of other atmospheric constituents. An example of water vapour is the mist that floats over a pot of boiling water.

<h3>What is Water vapor ?</h3>

The gaseous phase of water is known as water vapour, water vapour, or aqueous vapour. Within the hydrosphere, it is one type of water state. Water vapour can be created by the boiling or evaporation of liquid water as well as by the sublimation of ice. Like the majority of other atmospheric elements, water vapour is transparent.

Learn more about Water vapor here:

brainly.com/question/422363

#SPJ4

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A: the intensity

Explanation:

the closer the sound, the more intense it is. Think about the ambulance illustration in your text book (assuming you are using a physics textbook) : )

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The fluid in a grdulated cylinder should be read at the _____ of the meniscus.
galina1969 [7]
<span>The fluid in a graduated cylinder should be read at the BOTTOM of the meniscus.</span>
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An eagle flying at 35 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 10
Akimi4 [234]

Answer:

a)  F=475.7Hz

b)  F'=410.899Hz

Explanation:

From the question we are told that:

Velocity of eagle V_1=35m/s

Frequency of eagle F_1=440Hz

Velocity of Black bird V_2=10m/s

Speed of sound s=343m/s

a)

Generally the equation for Frequency is mathematically given by

 F=f_0(\frac{v-v_2}{v-v_1})

 F=440(\frac{343-10}{343-35})

 F=475.7Hz

b)

Generally the equation for Frequency is mathematically given by

 F'=f_0(\frac{v+v_2}{v+v_1})

 F'=440(\frac{343+10}{343+35})

 F'=410.899Hz

7 0
3 years ago
The balance Lenght of a potent ometer wire for a Cell of emf 1.62v is 90cm. if the Cell is replaced by another one of emf 1.08v.
andrew-mc [135]

Answer:

answer is 3.05v

Explanation:

hope it is helpful and briliant

6 0
2 years ago
A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori
olasank [31]

Explanation:

The given data is as follows.

        k = 130 N/m,       \Delta x = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = \frac{1}{2}kx^{2}

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}

   \frac{1}{2} \times 130 \times (0.17)^{2} = \frac{1}{2} \times 2.8 \times v^{2}

          v^{2} = \frac{3.757}{2.8}

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = \mu \times mg

                                    = 0.15 \times 2.8 \times 9.8

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           \frac{1}{2}mv^{2} = F_{f} \times d

           \frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d

             1.8515 = 4.116 \times d

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0
3 years ago
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