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Ivanshal [37]
2 years ago
10

Which of the following describes boundaries?

Engineering
1 answer:
torisob [31]2 years ago
7 0

Answer:

The correct option is;

(2) Boundaries are primarily used to indicate which functions a product will perform as opposed to the functions a user will perform

Explanation:

Product specifications accompanying most products, specifies the product  boundaries and limitations which are to be acceptable to (or meet the requirement of) the intended user

The intended user has the option to then set boundaries of use of the product within his possession.

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An electric field is expressed in rectangular coordinates by E = 6x2ax + 6y ay +4az V/m.Find:a) VMN if point M and N are specifi
Fittoniya [83]

Answer:

a.) -147V

b.) -120V

c.) 51V

Explanation:

a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).

b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.

c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.

Honestly, these things take practice to get used to. It's really hard to explain this.

3 0
2 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in
RideAnS [48]

To resolve this problem we have,

R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm

F_{f2} is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa

After that, we can calculate the flexural strenght for the square cross section using the previously value.

\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN

6 0
2 years ago
Fly thermostat, an automatic temperature controller for homes, learns the patterns for raising and lowering the temperature in a
Lubov Fominskaja [6]

Answer:

The Internet of things

Explanation:

When devices embedded with a network connectivity, software or sensors, are able to exchange data with the manufacturer through the network of physical things, then we say the Internet of things (IoT) has been defined.

Internet of things is when devices like smartphones, sensors and other smart devices are connected together which allows them to exchange data through the network of physical things. By exchanging data, these devices learn new patterns of adaptation.

In this case, the fly thermostat, learns the patterns for raising and lowering the temperature in a house through the Internet of things beacause the sensor has made it possible by automatically observing the temperature pattern of the house.

4 0
2 years ago
Read 2 more answers
Convert 273.15 mL at 166.0 mm of Hg to its new volume at standard pressure.​
zloy xaker [14]

Answer:

(166.0 mm Hg) (273.15 mL) = (760.0 mm Hg) (x)

4 0
2 years ago
Air enters a nozzle steadily at 280 kPa and 77°C with a velocity of 50 m/s and exits at 85 kPa and 320 m/s. The heat losses from
Arlecino [84]

The input values are the following

\left.T_{1}=350 K\right\ then

h_{1}=350.49 \frac{k j}{k g}, s_{1}=1.85708 \frac{K j}{K g \cdot K}

By using the energy equilibrium

\dot{E}_{i n}-\dot{E}_{o u t}=\Delta \dot{E}_{s y s t e m}=0 , \dot{E}_{i n}=\dot{E}_{o u t}

we haveT_{2}=297.2 K

eq (1)  \dot{m}\left(h_{1}+\frac{V_{1}^{2}}{2}\right)=\dot{m}\left(h_{2}+\frac{V_{2}^{2}}{2}\right)+\dot{Q}_{o u t} ∴

0=q+h_{2}-h_{1}+\frac{V_{2}^{2}-V_{1}^{2}}{2}

Now, for specific energy h2:

h_{2}=h_{1}-q_{o u t}-\frac{V_{2}^{2}-V_{1}^{2}}{2}

By replacing the eq (1) we have

h_{2}=350.49 \frac{k j}{k g}-3.2-\frac{\left(320 \frac{m}{s}\right)^{2}-\left(50 \frac{m}{m}\right)^{2}}{2}\left(\frac{1 \frac{k j}{k g}}{1000 \frac{m^{2}}{s^{2}}}\right)

h_{2}=297.34 \frac{k j}{k g}

By using a standard ideal-gas properties of air we have

T_{2}=297.2 K

6 0
2 years ago
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