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anyanavicka [17]
2 years ago
13

In 1923, the United States Army (there was no U.S. Air Force at that time) set a record for in-flight refueling of airplanes. Us

ing two refueling planes, an Airco DH-4B biplane was able to remain in flight for 37 hours. During the flight, the refueling planes were able to air-transfer a total of 687 gallons of fuel to the plane in 9 refueling transfers. Assume that the refueling nozzle had a diameter of 1.45 inches and each refueling took 2.31 minutes to perform. Calculate the velocity of the fuel through the nozzle. Assume that the fuel filled the entire cross-sectional area of the nozzle..

Physics
1 answer:
dybincka [34]2 years ago
6 0

Answer:

1.95m/s

Explanation:

Please view the attached file for the detailed solution.

The following were the conversion factors used in order to express all quatities in SI units:

1 gallon=0.00378541m^3\\1 inch=0.0254m\\1 minute=60s

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Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission
bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law                     P = σ A e T⁴

Wien displacement law   λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

       P₁ = σ A T₁⁴

T = 12000K

       P₂ = σ A T₂⁴

       P₂ / P₁ = T₂⁴ / T₁⁴

       P₂ / P₁ = (12000/3000)⁴

       P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

       λ T = 2,898 10-3

       λ₁ = 2.89810-3 / T

       λ₁ = 2,898 10-3 / 3000

       λ₁ = 0.966 10-6 m

      λ₁ = 966 nm

T = 12000K

      λ₂ = 2,898 10-3 / 12000

      λ₂ = 0.2415 10-6 m

      λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0
3 years ago
Please help I need this fast
Ghella [55]
The solution has reacted.
5 0
3 years ago
A stone is thrown with an initial speed of 12 m/s at an angle of 30o above the horizontal from the top edge of a cliff. If it ta
kherson [118]

Answer:

d=58m

Explanation:

From the question we are told that:

Initial Speed U=12m/s

Time T=5.6s

Angle \theta=30

Generally the  Newton's equation for motion is mathematically given by

d=d'+ut+\frac{at^2}{2}

d=12cos30*5.6

d=58m

8 0
2 years ago
Can i eat air? im hungry
Lerok [7]

Answer:

Yes

Explanation:

8 0
2 years ago
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4. How much milk at 5° C needs to be added to 250 g of coffee at 90° C to make the coffee drinkable at 60° C?
kenny6666 [7]

Answer:

dino :)

Explanation:

7 0
3 years ago
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