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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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Go, but only if the intersection is clear.

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Traffic at intersection can be complicated at times. If the green light comes on after a red light, you have the right of way to go, but you should be careful to only go when the intersection is clear to avoid an accident. Once using the road, a good driver should be conscious of the other road users, as accidents might happen from you claiming your-right of-way

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you know what sounds rlly nice? first have you seent he new corvettes? next u should here a nisan gtr. (like tanner foxes)

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The "divide and average" method, an old-time method for approximating the square root of any positive number a, can be formulate
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15. A cold-chamber die-casting machine operates automatically, supported by two industrial robots.The machine produces two zinc
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A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.11 m2 and whose thickn
Assoli18 [71]

Answer:

Explanation:

Given that,

The area of glass A_g = 0.11m^2

The thickness of the glass t_g=4mm=4\times10^-^3m

The area of the styrofoam A_s=11m^2

The thickness of the styrofoam t_s=0.20m

The thermal conductivity of the glass k_g=0.80J(s.m.C^o)

The thermal conductivity of the styrofoam  k_s=0.010J(s.m.C^o)

Inside and outside temperature difference is ΔT

The heat loss due to conduction in the window is

Q_g=\frac{k_gA_g\Delta T t}{t_g} \\\\=\frac{(0.8)(0.11)(\Delta T)t}{4.0\times 10^-^3}\\\\=(22\Delta Tt)j

The heat loss due to conduction in the wall is

Q_s=\frac{k_sA_s\Delta T t}{t_g} \\\\=\frac{(0.010)(11)(\Delta T)t}{0.20}\\\\=(0.55\Delta Tt)j

The net heat loss of the wall and the window is

Q=Q_g+Q_s\\\\=\frac{k_gA_g\Delta T t}{t_g}+\frac{k_sA_s\Delta T t}{t_g}\\\\=(22\Delta Tt)j +(0.55\Delta Tt)j \\\\=(22.55\Delta Tt)j

The percentage of heat lost by the window is

=\frac{Q_g}{Q}\times 100\\\\=\frac{22\Delta T t}{22.55\Delta T t}\times 100\\\\=97.6 \%

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3 years ago
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