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Ludmilka [50]
3 years ago
13

18. What do you call a change in state from a liquid to a gas?

Physics
2 answers:
makkiz [27]3 years ago
8 0
ANSWER: vaporization
EXPLANATION: Matter changes between the liquid and gas states through vaporization and condensation. Vaporization As liquid water is heated, its temperature rises. until it reaches 100°C. At this point, liquid water changes into water vapor. The change from a liquid to a gas is known as vaporization
Ket [755]3 years ago
5 0

Answer:

evaporation

Explanation:

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What changes into energy in the sun?
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The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom. During the fusion process, radiant energy is released. :D

4 0
3 years ago
Seismic waves do not travel along the Earth’s surface. Please select the best answer from the choices provided
mario62 [17]
<span>The correct answer to the question that is stated above is obviously letter b, false. 

</span>Seismic waves travel along the Earth’s surface.They are <span>waves of </span>energy<span> that travel through the </span>Earth<span>'s layers.  These waves are a result of an </span>earthquake <span>, </span>volcanic<span> eruptions, magma movement and large </span>landslides.
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3 years ago
Read 2 more answers
an asteroid flies close to the earth. gravity does what? A.repels the asteroid away from the earth. B. attracts the asteroid and
Marysya12 [62]
Due to the gravitational pull of earth the asteroid would be pulled towards earth, so the answer is B.
4 0
3 years ago
What point is labeled in the scatter plot? <br><br> A. (2,8)<br> B. (7,4)<br> C. (8,2)<br> D. (4,7)
Whitepunk [10]
D. Would be your answer
4 0
3 years ago
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If the range of the projectile is 4.3 m, the time-of-flight is T = 0.829 s, and air resistance is negligible, determine the foll
ankoles [38]

Answer

given,

range of the projectile = 4.3 m

time of flight = T = 0.829 s

v =\dfrac{d}{T}

v =\dfrac{4.3}{0.829}

     v = 5.19 m/s

vertical component of velocity of projectile

v_y = gt'

v_y = 9.8 \times {\dfrac{T}{2}}

v_y = 9.8 \times {\dfrac{0.829}{2}}

v_y =4.06\ m/s

a) Launch angle

 \theta = tan^{-1}(\dfrac{v_y}{v})

 \theta = tan^{-1}(\dfrac{4.06}{5.19})

    θ = 38°

b) initial speed of projectile

  v = \sqrt{v_x^2 + v_y^2}

  v = \sqrt{5.19^2 + 4.06^2}

         v = 6.59 m/s

c) maximum height reached by the projectile

     y_{max}=v_{avg}t'

     y_{max}=\dfrac{1}{2}v_y\dfrac{T}{2}

     y_{max}=\dfrac{1}{2}\times(g\dfrac{T}{2})\times\dfrac{T}{2}

     y_{max}=\dfrac{gT^2}{8}          

7 0
4 years ago
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