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3 years ago

Hi. I would like to know why one side of an island can get more rain (more rain forms), while the other gets less.

Engineering

2 answers:

adelina 88 [10]3 years ago

6 0

The elevation might be higher causing more rain

sdas [7]3 years ago

4 0

Answer:

It's because the moisture in the air condenses much quicker at a higher elevation. When the air condenses, rain clouds are formed. The side the air comes from is called the windward side, which is where there is more rainfall.

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In which type of shoot is continuous lighting used?

Fiesta28 [93]

A type of shoot in which continuous lighting used is: 1) studio.

<h3>What is a photoshoot?</h3>

A photoshoot simply refers to a photography session which typically involves the use of digital media equipment such as a camera, to take series of pictures (photographs) of models, group, things or places, etc., especially by a professional photographer.

<h3>The types of shoot.</h3>

Basically, there are four main type of shoot and these include the following:

Studio

Underwater

Action

Landscape

In photography, a type of shoot in which continuous lighting used is studio because it enhances the photographs.

Read more on photography here: brainly.com/question/24582274

#SPJ1

4 0

2 years ago

A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c

Eduardwww [97]

Answer:

$h=1.99998\ W/m^2.C$

$k=33.333\ W/m.C$

Explanation:

Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

$\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0$ Eq (1)

Where:

k is thermal Conductivity

$\dot e_{gen}$ is uniform thermal generation

$T(x) = a(L^2-x^2)+b$

$\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a$

Putt in Eq (1):

$-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C$

Energy balance is given by:

$Q_{convection}=Q_{conduction}$

$h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L$ Eq (2)

$T(x) = a(L^2-x^2)+b$

Putting x=L

$T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC$

$\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2$

From Eq (2)

$h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C$

7 0

3 years ago

In the dataset "RestRating", the 30 randomly selected patrons give the overall dining experience ratings (Outstanding, Very good

rusak2 [61]

Answer: incomplete question

Explanation:

6 0

3 years ago

Read 2 more answers

H2O enters a conical nozzle, operates at a steady state, at 2 MPa, 300 oC, with the inlet velocity 30 m/s and the mass flow rate

Colt1911 [192]

Answer:

The flow velocity at outlet is approximately 37.823 meters per second.

The inlet radius of the nozzle is approximately 0.258 meters.

Explanation:

A conical nozzle is a steady state device used to increase the velocity of a fluid at the expense of pressure. By First Law of Thermodynamics, we have the energy balance of the nozzle:

Energy Balance

$\dot m \cdot \left[\left(h_{in}+\frac{v_{in}^{2}}{2} \right)-\left(h_{out}+\frac{v_{out}^{2}}{2} \right)\right]= 0$ (1)

Where:

$\dot m$ - Mass flow, in kilograms per second.

$h_{in}$ , $h_{out}$ - Specific enthalpies at inlet and outlet, in kilojoules per second.

$v_{in}, v_{out}$ - Flow speed at inlet and outlet, in meters per second.

It is recommended to use water in the form of superheated steam to avoid the appearing of corrosion issues on the nozzle. From Property Charts of water we find the missing specific enthalpies:

Inlet (Superheated steam)

$p = 2000\,kPa$

$T = 300\,^{\circ}C$

$h_{in} = 3024.2\,\frac{kJ}{kg}$

$\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$

Where $\nu_{in}$ is the specific volume of water at inlet, in cubic meters per kilogram.

Outlet (Superheated steam)

$p = 600\,kPa$

$T = 160\,^{\circ}C$

$h_{out} = 2758.9\,\frac{kJ}{kg}$

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $h_{in} = 3024.2\,\frac{kJ}{kg}$ , $h_{out} = 2758.9\,\frac{kJ}{kg}$ and $v_{in} = 30\,\frac{m}{s}$ , then the flow speed at outlet is:

$35765-25\cdot v_{out}^{2} = 0$ (2)

$v_{out} \approx 37.823\,\frac{m}{s}$

The flow velocity at outlet is approximately 37.823 meters per second.

The mass flow is related to the inlet radius ( $r_{in}$ ), in meters, by this expression:

$\dot m = \frac{\pi \cdot v_{in}\cdot r_{in}^{2} }{\nu_{in}}$ (3)

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $v_{in} = 30\,\frac{m}{s}$ and $\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$ , then the inlet radius is:

$r_{in} = \sqrt{\frac{\dot m\cdot \nu_{in}}{\pi\cdot v_{in}}}$

$r_{in}\approx 0.258\,m$

The inlet radius of the nozzle is approximately 0.258 meters.

7 0

3 years ago

Describe how gene therapy can be a social issue and give rise to moral, ethical and legal debates. Justify your response in two

den301095 [7]

Answer: stigmatism and discrimination

Explanation:

Knowledge of genetic risks can lead to potential social and psychological consequences for the individual. Socially, knowledge from genetic tests may lead to stigmatization and discrimination within the community.

8 0

4 years ago