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Vera_Pavlovna [14]
3 years ago
5

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
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It is important to keeo a copy of your written plan and safety record s off-site. True or false
lyudmila [28]

Answer:

The answer for the question is true

Explanation:

If you get a virus or get hacked you will still have it saved

8 0
3 years ago
True or False: Stress can effectively be relieved through physical activity, getting enough rest and sleep, and relaxation techn
Neko [114]

Answer:

True

Explanation:

Actually this are some of the nitty gritty answers and ways to control or stip stress, lemme explain them you see stress most of the time may come from deep thoughts that are and are pushing you to the wall, and you in your state you as you react to that which is so demanding as it makes you tense,so if at you will incorporate physical activities like running, jogging or push ups the psychological tension in you is broken as you focus so much on the activities rather than the tension, then getting enough rest cools the mind and all your thoughts settle as in your brain starts to adapt to chilling and relaxation and enough sleep will actually make ones head to be at peace because if you lack enough sleep you might have an excruciating migraine when you are stressed up and finally relaxation techniques makes the body accept the situation and then you manouver out of it as you grow strongly.

Hope this will help!

7 0
3 years ago
Read 2 more answers
The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re
Anni [7]

Answer:

a) W_cycle = 200 KW , n_th = 33.33 %  , Irreversible

b) W_cycle = 600 KW , n_th = 100 %     , Impossible

c) W_cycle = 400 KW , n_th = 66.67 %  , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

  TL = 400 K

  TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

                                 W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

                                 n_th < n_max     ...... irreversible

                                 n_th = n_max     ...... reversible

                                 n_th > n_max     ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

                                n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And,                         n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 400 = 200 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 200 / 600 = 33.33 %

   - The type of process according to second Law of thermodynamics:

               n_th = 33.333 %                n_max = 66.67 %

                                       n_th < n_max  

      Hence,                Irreversible Process  

b) QH = 600 kW, QC = 0 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 0 = 600 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 600 / 600 = 100 %

   - The type of process according to second Law of thermodynamics:

                 n_th = 100 %                 n_max = 66.67 %

                                     n_th > n_max  

      Hence,               Impossible Process              

c) QH = 600 kW, QC = 200 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 200 = 400 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 400 / 600 = 66.67 %

   - The type of process according to second Law of thermodynamics:

               n_th = 66.67 %                 n_max = 66.67 %

                                     n_th = n_max  

      Hence,                Reversible Process

7 0
3 years ago
Use Excel, MatLab or a similar program to plot carrier thermal velocity as a function of temperature for both electrons and hole
Sidana [21]

Answer:

Solution 2:

The thermal energy of any particle is given as = \frac{1}{2}mv^{2}

where m= Effective mass of the particle,

v= Velocity of the particle

The average kinetic energy is given as =  \frac{3}{2}KT,

where K= Boltzmann Constant

T= Temperautre of the particle

At equilibrium,

1/2 mv² =  3/2 KT

Hence, v= \sqrt{\frac{3KT}{m}}

As, effective mass ratio is calculated with respect to rest mass of electron,

melectron = 0.11 * 9.11 *10-31 kg

mhole =  0.35 *9.11 *10-31 kg

K = 1.38 × 10-23 m2 kg s-2 K-1

Plotting, over a range of temperature of 0 K to 400 K, we obtain the attached Graphs (attached).

Solution 3:

The above two curves plotted are not identical as seen. From the same value of Temperature, and under identical conditons, the Thermal velocity solely depends on effective mass ratio. And it is inversely proportional to effective mass ratio. More the effective mass ratio, less the thermal velocity and flatter the slope of the curve and vice versa. As, the mass ratio of holes is more than that of electrons, the curve of electrons has a steeper slope than that of holes. Hence, the curves are not just identical.

8 0
3 years ago
Fresh cut potato strips with a moisture content of 50% (w/w) on wet basis are fried in peanut oil to produce French fries.
Rufina [12.5K]

Answer:

Percentage of oil uptake by the potato = 5.57% on a wet basis (including the moisture content)

Percentage of oil uptake by the potato = 7.24% on a dry basis (excluding the moisture content)

Explanation:

Starting with 1000kg of fresh potatoes,

It is given that this consists of 50% (w/w) of water.

Meaning that, 1000 kg of fresh potatoes consists of (1000 × 50%) of water = 500 kg of water.

If 1000 kg of potatoes consist of 500 kg of water, then the remaining 500 kg is pure potatoes (since it is stated that freshly cut potatoes have no oil content)

After frying, the weight of the French fries is 700 kg now.

But of this 700 kg, 23 % of its weight is moisture content, i.e. water,

23% of 700 kg = 161 kg.

This means that the amount of potatoes and oil in the French fries is 700 - 161 = 539 kg

But, recall, that the amount of potatoes in this process from the start is 500 kg. This amount doesn't change even on frying the fresh potatoes into French fries.

So, amount of oil in the French fries = 539 - 500 = 39 kg

Percentage of oil uptake by the potato = 100% × (amount of oil intake)/(total mass of potatoes) = 100% × 39/700 = 5.57% on a wet basis (including the moisture content)

And 100% × 39/539 = 7.24% on a dry basis (excluding the moisture content)

4 0
4 years ago
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