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Talja [164]
3 years ago
12

What are the weight restrictions for a small UAS, including everything onboard at the time

Engineering
1 answer:
Luden [163]3 years ago
5 0
B I think because GTOW is greater than 5 lbs than 55
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Technician A says when tightening lug or wheel nuts a torque wrench must be used. Technician B says that an air wrench can be us
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A, is your answer hope it helps
4 0
3 years ago
A 50000 N plane has wings with a span of 30 m and a chord of 6 m. How much cargo can this plane carry while cruising at 550 km/h
soldi70 [24.7K]

Answer:

The amount of cargo the plane can carry is 8707.89 N

Explanation:

The surface area of the wings facing the air = 30×6×2 × sin(2.5) = 15.7 m²

The speed of the plane 550 km/h = 152.78 m/s

The volume of air cut through per second = 15.7 × 152.78 = 2399.07 m³

The mass of air = Volume × Density = 2399.07 × 0.37 = 887.65 kg

Weight of air = Mass × Acceleration due to gravity = 887.65 × 9.81 = 8707.89 N

Given that the plane is already airborne, the additional cargo the plane can carry is given by the available lift force of the plane.

The amount of cargo the plane can carry = 8707.89 N

8 0
4 years ago
Who knows about welding ??
shtirl [24]
They joint together melting metal
5 0
4 years ago
Select the correct answer. Which if the following statements is an example of a good problem statement? to design a pair of runn
algol13

Answer:

To design a pair of running shoes for women.

Explanation:

A good problem statement will led a reader from a shared context to understanding of a problem and on to a proposed solution.

The elements of a good problem statement are ;

  • It should be addressing a gap in an idea
  • It should be vital enough to contribute to an existing body of research
  • It should offer room for further research
  • It should give itself to investigation through data collection.
5 0
3 years ago
A well-insulated, rigid vessel contains 3 kg of saturated liquid water at 40°C. The vessel also contains an electrical resistor
fenix001 [56]

Answer:

The final temperature in the vessel after the resistor has been operating for 30 min is 111.67°C

Explanation:

given information:

mass, m = 3 kg

initial temperature,  T₁ = 40°C

current, I = 10 A

voltage, V = 50 V

time, t = 30 min = 1800 s

Heat for the system because of the resistance is

Q = V I t

where

V = voltage (V)

I = current (A)

t = time (s)

Q = heat transfer to the system (J)

so,

Q = V x I x t

   = 50 x 10 x 1800

   = 900000

   = 9 x 10⁵ J

the heat transfer in the closed system is

Q = ΔU + W

where

U = internal energy

W = work done by the system

thus,

Q = ΔU + W

9 x 10⁵ = ΔU + 0, W = 0 because the tank is a well-insulated and rigid.

ΔU = 9 x 10⁵ J = 900 kJ

then, the energy change in the system is

ΔU = m c ΔT

ΔT = ΔU / m c, c = 4.186 J/g°C

     = 900 / (3 x 4.186)

     = 71.67°C

so,the final temperature (T₂)

ΔT = T₂ - T₁

T₂ = ΔT + T₁

    = 71.67°C + 40°C

    = 111.67°C

7 0
3 years ago
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