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1 year ago

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

1 answer:

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:

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Why would you expect the speed of light to be slightly less in the atmosphere then in a vacuum?

azamat

The speed of light to be slightly less in atmosphere then in vacuum because of absorption and re-emission of light by the atmospheric molecules occurred when light travels through a material

<u>Explanation:</u>

When light passes through atmosphere, it interacts or transmits through the transparent molecules in atmosphere. In this process of transmission through atmosphere, the light will be getting absorbed by them and some will get re-emitted or refracted depending upon wavelength.

But in vacuum the absence of any kind of particles will lead to no interaction and no energy loss, thus the speed of the light will be same in vacuum while due to interactions with molecules of atmosphere, there speed will be slightly less compared to in vacuum.

7 0

3 years ago

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0

2 years ago

Which of the following is a risk associated with taking performance-enchancing drugs.

Damm [24]

Answer:

The correct answer is All of the above.

Explanation:

7 0

2 years ago

An electron is confined to a one dimensional region, bounded by an infinite potential. If the energy of the electron in its firs

OLga [1]

Answer:

The energy in its ground state is 10 meV.

Explanation:

It is given that,

The energy of the electron in its first excited state is 40 meV.

Energy of the electron in any state is given by :

$E=\dfrac{n^2\pi^2h^2}{8mL^2}$

For ground state, n = 1

$E_1=\dfrac{\pi^2h^2}{8mL^2}$ .............(1)

For first excited state, n = 2

$40=\dfrac{2^2\pi^2h^2}{8mL^2}$ .............(2)

Dividing equation (1) and (2), we get :

$\dfrac{E_1}{40}=\dfrac{1}{4}$

$E_1=10\ meV$

So, the energy in its ground state is 10 meV. Hence, this is the required solution.

4 0

3 years ago

Object 1 of mass m moves with speed v in the positive direction. Object 2 of mass 3 m moves with speed 4 v in the negative x-dir

Klio2033 [76]

Answer:

This means that the kinetic energy of second object is 48times that of the first object

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion e.g motion of an accelerating car. Mathematically,

Kinetic energy = 1/2mv² where;

m is the mass of the object

v is the velocity of the object

If Object 1 of mass m moves with speed v in the positive direction, its kinetic energy will be expressed as;

K1 = 1/2mv²

For Object 2 of mass 3m moving with speed 4v in the negative x-direction, its kinetic energy can be expressed as;

K2 = 1/2(3m)(4v)²

K2 = 1/2(3m)(16v²)

K2 = (3m)(8v²)

K2 = 24mv²

To compare the kinetic energy of both bodies, we will take the ratio of K2:K1 to have;

K2/K1 = 24mv²/(1/2)mv²

K2/K1 = 24/(1/2)

K2/K1 = 48

K2 = 48K1

This means that the kinetic energy of second object is 48times that of the first object and moving in the negative x direction since the body of mass 3m initially moves in the negative x direction.

3 0

3 years ago