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3 years ago

11

The article provides information by using a list. What does it list? A. Thanksgiving food B. places where clams can be found C.

birds served at the first Thanksgiving D. four men who went on a "fowling" mission

1 answer:

Gelneren [198K]3 years ago

7 0

Answer:

C

Explanation:

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Some Tiny College staff employees i s are information technology (IT) personnel. Some IT personnel provide technology support fo

Westkost [7]

Answer:

solution in the picture attached

Explanation:

3 0

3 years ago

Ensure at least ___ distance around fire sprinkler heads, safety showers, eyewash units, and heating and cooling units to ensure

vampirchik [111]

90 inches

Explanation:

According to OSHA requirement, the distance around safety showers and eyewash should be between 82-96 inches off the flow. This will allow for maximum diameter of spray.

Learn More

Safety distance around safety showers:brainly.com/question/11123362

Keywords: distance, fire sprinkler head, safety showers, eyewash units,heating and cooling units

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3 years ago

The advantage of an interferometer is that

victus00 [196]

It can provide measurements of stars with a higher angular resolution than is possible with conventional telescopes.

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1 year ago

A building window pane that is 1.44 m high and 0.96 m wide is separated from the ambient air by a storm window of the same heigh

sdas [7]

Answer:

the rate of heat loss by convection across the air space = 82.53 W

Explanation:

The film temperature

$T_f = \frac{T_1+T_2}{2} \\\\= \frac{20-10}{2}\\\\= \frac{10}{2}\\\\= 5^0\ C$

to kelvin = (5 + 273)K = 278 K

From the " thermophysical properties of gases at atmospheric pressure" table; At $T_f$ = 278 K ; by interpolation; we have the following

$\frac{278-250}{300-250}= \frac{v-11.44(10^{-6})}{15.89(10^{-6})-11.44(10^{-6})}$ → v 13.93 (10⁻⁶) m²/s

$\frac{278-250}{300-250}= \frac{k-22.3(10^{-3}}{26.3(10^{-3}-22.3(10^{-3})}$ → k = 0.0245 W/m.K

$\frac{278-250}{300-250}= \frac{\alpha - 15.9(10^{-6})}{22.5(10^{-6}-15.9(10^{-6})}$ → ∝ = 19.6(10⁻⁶)m²/s

$\frac{278-250}{300-250}= \frac{Pr-0.720}{0.707-0.720}$ → Pr = 0.713

$\beta = \frac{1}{T_f} \\=\frac{1}{278} \\ \\ = 0.00360 \ K ^{-1}$

The Rayleigh number for vertical cavity

$Ra_L = \frac{g \beta (T_1-T_2)L^3}{\alpha v}$

= $\frac{9.81*0.00360(20-(-10))*0.06^3}{19.6(10^{-6})*13.93(10^{-6})}$

= $8.38*10^5$

$\frac{H}{L}= \frac{1.44}{0.06} \\ \\= 24$

For the rectangular cavity enclosure , the Nusselt number empirical correlation:

$Nu_L = 0.42(8.38*10^5)^{\frac{1}{4}}(0.713)^{0.012}(24){-0.3}$

$NU_L= \frac{hL}{k}= 4.878$

$\frac{hL}{k}= 4.878$

$\frac{h*0.06}{0.0245}= 4.878$

$h = \frac{4.878*0.0245}{0.06}$

h = 1.99 W/m².K

Finally; the rate of heat loss by convection across the air space;

q = hA(T₁ - T₂)

q = 1.99(1.4*0.96)(20-(-10))

q = 82.53 W

3 0

3 years ago

Calculate the conductance of a 4.7 ohm resistor

wolverine [178]

Answer:

0.213 S

Explanation:

To calculate the conductance of an electrical component, in this case, a resistor, simply divide 4.7 into 1.

Conductance = 1 / 4.7 ohms

Conductance = 0.213 S

Cheers.

3 0

3 years ago