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ivanzaharov [21]

3 years ago

14

In ________ programming, the programming is centered on objects that are created from abstract data types that encapsulate data

and functions together.

1 answer:

Contact [7]3 years ago

8 0

Answer:

<em>Object-oriented</em>

Explanation:

<em>Object Oriented programming</em> <em>(OOP)</em> is a specific way of programming, where the code is organized in units called classes, from which objects are created that are related to each other to achieve the objectives of the applications. Object-oriented programming took over as the dominant programming style in the mid-1980s, largely due to the influence of C ++. Its dominance was consolidated thanks to the rise of graphical user interfaces, for which object-oriented programming is particularly well suited. Its most important characteristics are the following:

Encapsulation

Polymorphism

Abstraction

Inheritance

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Answer:

The correct option is;

D. She will need to research the problem even more

Explanation:

In other to properly design the pool umbrella that will be reliable and meet the needs of the particular customer, she will need to carry out a basic research about the environment where the pool umbrella is to be used, the presence of other factors that can impact on an idea for the pool umbrella.

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4 years ago

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How do heat and our use of electricity affect our daily activities and the environment

Andrews [41]

Answer:

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2 years ago

Risks are Not Perceived Differently from What is Happening<br> False<br> True

Gnoma [55]

Answer:

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3 years ago

An astronaut weighs 125 lbs on the surface of the Earth. What would she weigh in lbs if she is resting in the space shuttle that

alina1380 [7]

Answer:

weight of astronaut $F_2 = 113.38 lb$

Explanation:

Given data:

astronaut weight is = 125 lbs

distance of shuttle from surface of earth is 200 mils

we know that $F = G \frac{m_1 m_2}{r^2}$

where G, m_1, m_2 are constant value, so we have

$\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$

radius of earth is 4000 miles

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3 0

3 years ago

The heat flux through a 1-mm thick layer of skin is 1.05 x 104 W/m2. The temperature at the inside surface is 37°C and the tempe

miss Akunina [59]

Answer:

a) Thermal conductivity of skin: $k_{skin}=1.5W/mK$

b) Temperature of interface: $T_{interface}=35.6\°C$

Heat flux through skin: $\frac{Q}{A}=2100W/m^2$

Explanation:

a)

$k=\frac{QL}{A(T_{2}-T_{1})}$

Where: $k$ is thermal conductivity of a material, $\frac{Q}{A}$ is heat flux through a material, $L$ is the thickness of the material, $T_{1}$ is the temperature on the first side and $T_{2}$ is the temperature on the second side

$k_{skin}=\frac{QL}{A(T_{2}-T_{1})}$

$k_{skin}=\frac{Q}{A}*\frac{L}{(T_{2}-T_{1})}$

$k_{skin}=1.05*10^{4}*\frac{1*10^{-3}}{(37-30)}$

$k_{skin}=1.5W/mK$

b)

$k_{insulation}=\frac{k_{skin}}{2}$

$k_{insulation}=\frac{1.5}{2}$

$k_{insulation}=0.75W/mK$

The heat flux between both surfaces is constant, assuming the temperature is maintained at each surface.

$\frac{Q}{A}=\frac{k(T_{2}-T_{1})}{L}$

$\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}=\frac{k_{insulation}(T_{interface}-T_{insulation})}{L_{insulation}}$

$\frac{1.5*(37-T_{interface})}{0.001}=\frac{0.75*(T_{interface}-30)}{0.002}$

$55500-1500T_{interface}=375T_{interface}-11250$

$1875T_{interface}=66750$

$T_{interface}=35.6\°C$

$\frac{Q}{A}=\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}$

$\frac{Q}{A}=\frac{1.5*(37-35.6)}{0.001}$

$\frac{Q}{A}=2100W/m^2$

3 0

3 years ago