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2 years ago

NASA SPACE SHUTTLE QUESTION:

Engineering

1 answer:

vfiekz [6]2 years ago

8 0

Explanation:

The Challenger accident made them aware of the risks. They had been a little naive, since it was never believed that such a thing could happen. This, together with Space Shuttle Columbia disaster revealed the risks of the space shuttle program, which also had very high maintenance costs.

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Typically each development platform consists of the following components, except:Select one:a.Operating systemb.System softwarec

damaskus [11]

Typically each development platform consists of the following components except compilers and assemblers

The platform development simply means the development of the fundamental software which is vital in making hardware work.

Operating system: This refers to the low-level software that communicates with the hardware so that other programs can be able to run.

System software: This is the software that's designed in order to provide a platform for the other software. Examples include search engines, Microsoft Windows, etc.

Compilers and assemblers: Compliers are sued in converting source code to a machine-level language. Assembler is used in converting assembly code to machine code.

Hardware platform: This is a set of hardware where the software applications are run.

In conclusion, the correct option is Compilers and assemblers.

Read related link on:

brainly.com/question/21650058

4 0

2 years ago

Why is the uniaxial tension test commonly used to evaluate the mechanical properties of metals?

inn [45]

Answer:

Because with this test you can determine complex material parameters like Young’s modulus, yield strength, ultimate strength and elongation at break. This is important because it provides us with the factor of safety that needs to be built-in the products using these materials.

Explanation:

This test consists in place the material between to tweezers to subdued the material into a stress-strain test. The figure shows the procedure.

In the x axe the strain and the y axe the stress.

3 0

3 years ago

Thick fluids such as asphalt and waxes and the pipes in which they flow are often heated in order to reduce the viscosity of the

Gala2k [10]

Answer:

what id the answer

Explanation:

7 0

2 years ago

Consider a mixing tank with a volume of 4 m3. Glycerinflows into a mixing tank through pipe A with an average velocity of 6 m/s,

Svetach [21]

This question is incomplete, the complete question as well as the missing diagram is uploaded below;

Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.

Take $p_o$ = 880 kg/m³ and $p_{glycerol$ = 1260 kg/m³

Answer:

the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³

Explanation:

Given that;

Inlet velocity of Glycerin, $V_A$ = 6 m/s

Inlet velocity of oil, $V_B$ = 3 m/s

Density velocity of glycerin, $p_{glycerol$ = 1260 kg/m³

Density velocity of glycerin, Take $p_o$ = 880 kg/m³

Volume of tank V = 4 m

from the diagram;

Diameter of glycerin pipe, $d_A$ = 100 mm = 0.1 m

Diameter of oil pipe, $d_B$ = 80 mm = 0.08 m

Diameter of outlet pipe $d_C$ = 120 mm = 0.12 m

Now, Appling the discharge flow equation;

$Q_A + Q_B = Q_C$

$A_Av_A + A_Bv_B = A_Cv_C$

π/4 × ( $d_A$ )² $v_A$ + π/4 × ( $d_B$ )² $v_B$ = π/4 × ( $d_C$ )² $v_C$

we substitute

π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)² $v_C$

0.04712 + 0.0150796 = 0.0113097 $v_C$

0.0621996 = 0.0113097 $v_C$

$v_C$ = 0.0621996 / 0.0113097

$v_C$ = 5.5 m/s

Now we apply the mass flow rate condition

$m_A + m_B = m_C$

$p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C$

so we substitute

1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5

1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335

59.3712 + 13.27 = 0.06220335p

72.6412 = 0.06220335p

p = 72.6412 / 0.06220335

p = 1167.8 kg/m³

Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³

4 0

3 years ago

A viscous fluid flows in a 0.10-m-diameter pipe such that its velocity measured 0.012 m away from the pipe wall is 0.8 m/s. If t

maksim [4K]

Answer:

A) centerline velocity = 1.894 m/s

B) flow rate = 7.44 x 10^(-3) m³/s

Explanation:

A) The flow velocity intensity for the input radial coordinate "r" is given by;

U(r) = (Δp•D²/16μL) [1 - (2r/D)²]

Velocity at the centre of the tube can be expressed as;

V_c = (Δp•D²/16μL)

Thus,

U(r) = (V_c)[1 - (2r/D)²]

From question, diameter = 0.1m,thus radius (r) = 0.1/2 = 0.05m

But we are to find the velocity at the centre of the tube, thus;

We will use the radius across the horizontal distance which will be;

0.05 - 0.012 = 0.038m

Thus, let's put 0.038 for r in the velocity intensity equation and put other relevant values to get the velocity at the centre.

Thus;

U(r) = (V_c)[1 - (2r/D)²]

0.8 = (V_c)[1 - {(2 * 0.038)/0.1}²]

0.8 = (V_c)[1 - (0.76)²]

V_c = 0.8/0.4224 = 1.894 m/s

B) flow rate is given by;

ΔV = Average Velocity x Area

Now, average velocity = V_c/2

Thus, average velocity = 1.894/2 = 0.947 m/s

Area(A) = πr² = π x 0.05² = 0.007854 m²

So, flow rate = 0.947 x 0.007854 = 7.44 x 10^(-3) m³/s

4 0

3 years ago