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prisoha [69]
4 years ago
10

Why is flexibility the most obvious benefit of road transportation select all that apply

Engineering
1 answer:
ser-zykov [4K]4 years ago
7 0
What am I going to select?? What are my choices bro????
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Problem #1: A centrifugal compressor has a mass flow rate of 1.6 kg/s while rotating at 11,000 rpm. The impellor is a radial imp
saveliy_v [14]

Answer:

a)  The slip coefficient is 0.9

b) Blade tip speed is 345.57 m/s

c) Stagnation temperature exiting the impellor is 416.84 k

d) Exit flow velocity is 84.88 m/s

e) Exit flow angle is 76.2°

f) Exit static temperature is 353.84 k

g) Impellor exit Mach number is 0.943

Explanation:

Flow at entry is axial α₁ = 0

Tagential velocity at entry V_{w1}= 0

Blade at exit is radial  β₂ = 0

μ₂ = V_{w2}

3 0
3 years ago
A 15-kg iron block initially at 280°C is quenched in an insulated tank that contains 100 kg of water at 18? Assuming the water t
Serhud [2]

Answer:

Explanation: Reverse card.

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4 years ago
It is proposed to deposit a 5 μm thick nickel coating uniformly on all surfaces of a ceramic strip measuring 15 cm x 5 cm x 2 cm
MatroZZZ [7]

Answer:

Check the explanation

Explanation:

Let’s take for instance, when an object with a mass of 10 kg (m = 10 kg) is moving at a 5 meters per second (v = 5 m/s) velocity rate, the kinetic energy is equal to 125 Joules

Kindly check the attached images below to get the step by step explanation to the question above.

8 0
3 years ago
I dont undertand this coding problem (Java):
chubhunter [2.5K]

Check The Attachment.

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6 0
4 years ago
External crack of length of 3.0 mm was detected on the surface of the shaft of wind turbine made from 4340 steel. The diameter o
sveticcg [70]

Answer:

The correct answer is "K_c=6.0369 \ MPa\sqrt{m}".

Explanation:

Given:

Maximum load,

P = 50,000 N

Crack length,

a = 3mm

or,

  = 3×10⁻³ m

Diameter,

d = 32 mm

As we know,

⇒  Maximum stress, \sigma=\frac{P}{A}

                                      =\frac{50000}{(\frac{\pi}{4}\times 32^2)}

                                      =62.20 \ N/mm^2

Now,

⇒  Fracture tougness, K_c=Y \sigma\sqrt{\pi a}

On substituting the values, we get

                                           =1\times 62.20\times \sqrt{3.14\times 3\times 10^{-3}}

                                           =6.0369 \ MPa\sqrt{m}

4 0
3 years ago
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