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3 years ago

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What is the heat flux (W/m2) to an object when subjected to convection heat transfer environment given: 24 °C = the surface temp

erature of the object, 82 °C = the bulk temperature of the fluid adjacent to the object, and the convection heat transfer coefficient = 49 W/(m^2 K)?

1 answer:

xxMikexx [17]3 years ago

4 0

Answer:

attached below

Explanation:

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What are the most used electronic tools for electronic works?

Zanzabum

Answer:

Needle-nose pliers.

Hemostats.

A spool of 30-gauge wire.

A spool of solder wick.

A spool of solder.

A loupe.

Three sets of tweezer

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3 years ago

You just purchased a 400-L rigid tank for a client who works in the gas industry. The tank is delivered pre-filled with 3 kg of

solniwko [45]

Answer:

the pressure reading when connected a pressure gauge is 543.44 kPa

Explanation:

Given data

tank volume (V) = 400 L i.e 0.4 m³

temperature (T) = 25°C i.e. 25°C + 273 = 298 K

air mass (m) = 3 kg

atmospheric pressure = 98 kPa

To find out

pressure reading

Solution

we have find out pressure reading by gauge pressure

i.e. gauge pressure = absolute pressure - atmospheric pressure

first we find absolute pressure (p) by the ideal gas condition

i.e pV = mRT

p = mRT / V

p = ( 3 × 0.287 × 298 ) / 0.4

p = 641.44 kPa

so

gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 641.44 - 98

gauge pressure = 543.44 kPa

6 0

3 years ago

The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content

MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

$Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf$

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

$\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}$

Therefore, $V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}$

$V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy$

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since $1 yd^{3}= 27 ft^{3}$

The embankment requires water of 6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

$Unit_{weight}=\frac {17451720}{498594}=35.00186 lb$

1 gallon is approximately $8.35 yd^{3}$ hence

$\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}$

That's approximately 4.2 gallons

7 0

3 years ago

A strain gauge with a 5 mm gauge length gives a displacement reading of 1.25 um. Calculate the stress value given by this displa

KengaRu [80]

Answer:

stress = 50MPa

Explanation:

given data:

Length of strain guage is 5mm

displacement $\delta = 1.25 \mu m =\frac{1.25}{1000} = 0.00125 mm$

stress due to displacement in structural steel can be determined by using following relation

$E =\frac{stress}{strain}$

$stress = E \times strain$

where E is young's modulus of elasticity

E for steel is 200 GPa

$stress = 200\times 10^3 *\frac{1.25*10^{-3}}{5}$

stress = 50MPa

7 0

4 years ago

Suppose a student rubs a Teflon rod with wool and then briefly touches it to an initially neutral aluminum rod suspended by insu

Oliga [24]

Answer:

Due to touch of teflon, its charge will reduce but will not go to zero. Some amount of its initial charge will be transferred to Aluminum rod. So, aluminum rod will have a non-zero negative charge.

Explanation:

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4 years ago