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Ivahew [28]
3 years ago
11

As an engineer who has just finished taking engineering materials course, your first task is to investigate the causes of an aut

omobile accident. Your findings show that the right rear wheel has broken off at the axle. The axle is bent. The fracture surface reveals a Chevron pattern pointing toward the surface of the axle. Suggest a possible cause for the fracture and why?
Engineering
1 answer:
SIZIF [17.4K]3 years ago
3 0

Answer is given below

Explanation:

  • Evidence shows that the axle was not broken before the accident, while the clumsy axle meant that the wheel was still attached when the load was applied.
  • This indicates that the Chevron prototype wheel suffered a severe impact shock, which caused the failure of the transmission to the axle. Preliminary evidence suggests that the driver lost control and crashed.
  • Further examination of the surface, microstructure and structure and characteristics of the fracture can be modified if the axle is properly prepared
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Answer:

The circuit is shown in my attachments, please take a good look

3 0
3 years ago
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Our rule-of-thumb for presenting final results is to round to three significant digits or four if the first digit is a one. By t
olga_2 [115]

Answer:

To four significant digits = 2097 psi

Explanation:

<u>Applying the rule of thumb </u>

σ = Mc/I  ---- ( 1 )

M = 1835 Ibf in ,  I/c = 0.875 in^3

∴ c/l = 1 / 0.875 = 1.1429

back to equation 1

σ = 1835 * 1.1429 = 2097.2215 psi

To four significant digits = 2097 psi

4 0
3 years ago
Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horiz
Naddik [55]

Answer:

Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s       and ρ= 850 kg/m³  

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ =  ρv

   =850 x 0.00062

   = 0.527 kg/m·s  

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

Q = \dfrac{\Delta P \pi D^4}{128 \mu L}

Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}

Q = 5.06 x 10⁻⁸ m³/s

4 0
3 years ago
A binary star system consists of two stars of masses m1m1m_1 and m2m2m_2. The stars, which gravitationally attract each other, r
d1i1m1o1n [39]

Answer:

          a_c_2=\dfrac{a_c_1\times m_1}{m_2}

Explanation:

The question is: <em>Find the magnitude of the centripetal acceleration of the star with mass m₂</em>

The <em>centripetal acceleration</em> is the quotient of the centripetal force and the mass.

                a_c=\dfrac{F_c}{m}

Thus, you can write the equations for each star:

     

       a_c_1=\dfrac{F_c_1}{m_1}

       a_c_2=\dfrac{F_c_2}{m_2}

As per Newton's third law, the centripetal forces are equal in magnitude. Then:

       a_c_1\times m_1=a_c_2\times m_2

Now you can clear a_c_2:

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6 0
4 years ago
A fluid flows along the x axis with a velocity given by V = (xt) i ˆ, where x is in feet and t in seconds. (a) Plot the speed fo
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Answer

See explanation for step by step procedures towards getting answers

Explanation:

Given that;

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See attachments for further explanations

7 0
3 years ago
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