All categories

3 years ago

Select the material used to clean a prototype board before soldering.

Engineering

1 answer:

Dvinal [7]3 years ago

3 0

You need to use a cotton ball because it’s gets it clean and it’s not wet like the water or cleaner fluid

You might be interested in

Danny enjoys studying transportation networks and systems. Which field of engineering should he pursue?

andre [41]

Answer:

Civil Engineering.

Explanation:

Civil Engineering is a field of engineering study that deals with professional practising in designing and developing infrastructures such as transport system and networks, buildings, etc.

Civil engineering is regarded as the second--oldest form of engineering studies in the field of engineering.

A person who studies civil engineering is able to design and develop major projects in transportations system such as road construction, bridges, airports, tunnels, etc.

So, Danny, who is interested in studying transporation networks and system should opt for studying civil engineering.

8 0

3 years ago

How do you know when an equation is (In)

pav-90 [236]

If a vertical line crosses the relation on the graph only once in all locations, the relation is a function. However, if a vertical line crosses the relation more than once, the relation is not a function. Using the vertical line test, all lines except for vertical lines are functions

3 0

3 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

8 0

4 years ago

Please read the short case at the end of Chapter 8, "Mary Barra of General Motors Values Culture". Use the following prompts to

Tom [10]

Answer: Incoherent question

Explanation: This is an act of plagiarism at subjecting the tutor to unnecessary stress at answering the purported question.

4 0

4 years ago

Consider an ideal gas undergoing a constant pressure process from state 1 to state

Radda [10]

Answer:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy at constant pressure:

$ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}$

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

$\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,$

We obtain the expression to compute the specific entropy change:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Best regards.

6 0

4 years ago