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aivan3 [116]
3 years ago
15

Water changes from a liquid to a gas in the process of evaporation. Please select the best answer from the choices provided T F

Physics
2 answers:
Inessa05 [86]3 years ago
7 0

This is true, by the process of evaporation the liquid, changes to another type of matter that in this case is gas, this happens because the molecules move and vibrate so quickly that they escape into the atmosphere as molecules of water vapor.

const2013 [10]3 years ago
6 0
The correct answer is T.
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A woman has a mass of 55kg on earth, what is her mass on the moon​
KengaRu [80]

Answer:

55 kg

Explanation:

Mass does not change and is not dependant on gravity. So even though your weight will be less on the moon your mass won't change.

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Wave-particle duality tells us that wave and particle models apply to all objects whatever the size, so why don't we observe wav
Genrish500 [490]

Answer:

Because the wavelengths of macroscopic objects are too short for them to be detectable.

Explanation:

Wavelength of an object is given by de Broglie wavelength as:

\lambda=\frac{h}{mv}

Where, 'h' is Planck's constant, 'm' is mass of object and 'v' is its velocity.

So, for macroscopic objects, the mass is very large compared to microscopic objects. As we can observe from the above formula, there is an inverse relationship between the mass and wavelength of the object.

So, for vary larger masses, the wavelength would be too short and one will find it undetectable. Therefore, we don't observe wave properties in macroscopic objects.

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3 years ago
Question 2 (10 points)
Simora [160]

Answer:

traping

Explanation:

8 0
3 years ago
Read 2 more answers
If the temperature is 50 degrees and the dew point is 50 degrees, will it rain?
Paul [167]
There is a possibility but not extremely likely
4 0
3 years ago
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For anti-ballistic missile system, the time of flight tf is determined by the initial speed v0 of the missile and the maximum ra
umka2103 [35]
Refer to the diagram shown below.

The following discussion assumes a simplistic analysis that ignores air resistance and variations in the terrain that the missile travels over.

Let the launch velocity be V₀ at an angle of θ relative to the horizontal.
The horizontal component of velocity is V₀ cosθ.
If the time of flight is t_{f}, then
r=V_{o} \, t_{f}
where r = the range of the missile.

Also, the time, t, when the missile is at ground level is given by
0=V_{o} sin\theta \, t- \frac{1}{2}gt^{2}
where g = acceleration due to gravity.

t = 0 corresponds to when the missile is launched. Therefore
t_{f} =  \frac{2V_{o}sin\theta}{g}

Therefore
r= \frac{2V_{o}^{2} sin\theta cos\theta}{g} = \frac{V_{o}^{2} sin(2\theta)}{g}

Typically, θ=45° to achieve maximum range, so that
r= \frac{V_{o}^{2}}{g}

This analysis is more applicable to a scud missile rather than a powered, guided missile.

Answer:
t_{f} =  \frac{r}{V_{o} cos\theta} \\\\ r= \frac{V_{o}^{2} sin(2\theta)}{g}
Usually, θ=45°

6 0
3 years ago
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