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elena-14-01-66 [18.8K]
3 years ago
10

What layer of the atmosphere contains all of the weather and thus the most water vapor?

Physics
1 answer:
Veronika [31]3 years ago
5 0

The troposphere is the lowermost layer of the Earth's atmosphere. Most of the weather phenomena, systems, convection, turbulence and clouds occur in this layer, although some may extend into the lower portion of the stratosphere.

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When heating water, during what temperature range will the temperature cease to change for some time?
Morgarella [4.7K]

Answer: Option (B) is the correct answer.

Explanation:

As we know that the temperature when the vapor pressure of liquid becomes equal to the atmospheric pressure surrounding the liquid. And, during this temperature liquid state of substance changes into vapor state.

But during this process of change in state of substance the temperature will cease to change for some time because unless and until all the liquid molecules do not convert into vapor state the temperature will not rise or change.

As the boiling point of water is 100^{o}C so the temperature ceases to change from 98^{o}C to 102^{o}C.

Therefore, we can conclude that when heating water, during 98^{o}C to 102^{o}C temperature range the temperature will cease to change for some time.

7 0
3 years ago
Find an expression for the electric field E⃗ at the center of the semicircle. Hint: A small piece of arc length Δs spans a small
Hatshy [7]

Answer:

Electric Field a the centreE=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)

Explanation:

<u>Given:</u>

Total charge on the semicircle =Q

Radius of the semicircle=R

Let consider a elemental charge on the semicircle at an angle \theta\\ with the horizontal. Since the The charge distribution is symmetrical about y axis so the there will symmetrical charge on other side. The horizontal component will cancel out each other and vertical component will get added so let dE be the electric Field due to elemental charge

Let \lambda be the charge per unit length such that\lambda=\dfrac{Q}{\pi R}

k=\dfrac{1}{4\pi \epsilon_0}

Total Electric Field at the centre

=2dE\sin\theta\\=2\dfrac{k\lambda }{R}\int \sin \theta d\theta\\

integrating 0 to \dfrac{\pi}{2}

E=\dfrac{2k\lambda}{R}(-\vec j)

E=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)

So the Electric field at the centre is calculated.

7 0
3 years ago
(Serway 9th ed., 6-27) The mass of a sports car is 1200 kg. The shape of the body is such that the aerodynamic drag coefficient
kkurt [141]

Answer:

a = - 0.248 m/s²

Explanation:

Frictional drag force

F = ½ *(ρ* v² * A * α)

ρ = density of air  , ρ = 1.295 kg/m^3

α = drag coef , α = 0.250

v = 100 km/h x 1000m / 3600s

v =  27.77 m/s

A = 2.20m^2

So replacing numeric in the initial equation

F = ½ (1.295kg/m^3)(27.77m/s)²(2.30m^2)(0.26)

F = 298.6 N

Now knowing the force can find the acceleration

a = - F / m

a = - 298.6 N / 1200 kg

a = - 0.248 m/s²

3 0
3 years ago
Name six processes that are often involved in scientific inquiry?
lukranit [14]
1.identification of the problem.2. observation.3.formulation of the hypothesis.4. experimentation.5. analysis of data.6. drawing your conclusion.
4 0
3 years ago
Read 2 more answers
How high (in meters) can a 40 N force move a load, when 395 J of work is done? (round to the nearest 10th)
alex41 [277]

Answer:

9.9 m

Explanation:

From the question given above, the following data were obtained:

Force (F) = 40 N

Workdone (Wd) = 395 J

Height (h) =?

The height can be obtained as follow:

Potential energy = mgh = Fh

395 = 40 × h

Divide both side by 40

h = 395 / 40

h = 9.9 m

Thus, the height is 9.9 m.

7 0
3 years ago
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