3.) Technician A says that a scan tool can be used to verify engine operating temperature,
Technician B says that a refractometer can be used to verify engine operating temperature.
Who is right?
OA. A only
OB. B only
OC. Both A and B
OD. Neither A nor B

Answer:

50421.6 m³

Explanation:

The river has an average rate of water flow of 59.6 m³/s.

Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:

Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s

The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken

time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds

The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³

Answer:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

Explanation:

Given

[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,[/tex]

[tex]0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,[/tex]

[tex]0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,[/tex]

[tex]0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.[/tex]

Required

Plot a steam and leaf display for the given data

Start by categorizing the data by their tenth values:

[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.[/tex]

[tex]0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,[/tex]

[tex]0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.[/tex]

[tex]0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.[/tex]

[tex]0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.[/tex]

[tex]0.78.[/tex]

The 0.3's is will be plotted as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]

The 0.4's is as follows:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}[/tex]

The 0.5's is as follows:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}[/tex]

The 0.6's is as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]

Lastly, the 0.7's is as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

The combined steam and leaf plot is:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

Answer:

a) the friction angle of the sand is 36.87°

b) the deviator stress at failure is 450 kN/m³

Explanation:

Given the data in the question;

For a consolidated drained test

The effective major principle stresses

σ₃ = σ₃' = 80 kN/m²

and

σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 80 kN/m² + 240 kN/m²  = 320 kN/m²

now

a) friction angle of the sand

σ₁' = σ₃'tan²( 45° + Ф/2' ) + 2c' tan( 45° + Ф/2 )

for sand; c' = 0

so

σ₁' = σ₃'tan²( 45° + Ф/2' )

we substitute

320 = 80 tan²( 45° + Ф/2' )

Ф' = 2 × [ tan⁻¹ (√[tex]\frac{320}{80}[/tex]) - 45° ]

Ф' = 2 × [ 63.4349° - 45° ]

Ф' = 2 × 18.4349

Ф' = 36.87°

Therefore,  the friction angle of the sand is 36.87°

b)  deviator stress

σ₁' = σ₃'tan²( 45° + Ф/2' )

σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )

σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )   3.597

150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 150tan²( 45° + 36.87°/2 )

150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600

(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600 - 150

(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex]  = 450 kN/m³

Therefore, the deviator stress at failure is 450 kN/m³

Answer:

a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b) length of the specimen is 66.97 mm

Explanation:

Given the data in the question;

a) Determine the maximum stress that can be applied without plastic deformation

when know that; maximum stress σ[tex]_{max}[/tex]  = F / A

where F is the force in the rod ( 39872 N )

A is the cross-sectional area of the rod ( 100 mm² )

so we substitute;

σ[tex]_{max}[/tex]  = 39872 N / 100 mm²

σ[tex]_{max}[/tex]  = 398.72 N/mm²

Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b)  

strain in the members can be calculated using the expression

ε = σ / E

where σ is the stress in the rod

E is the module of elasticity (  110 GPa = 110000 N/mm² )

(Sl-L) / L = σ/E

where Sl-L is the change in length of the member

L is the original length of the specimen

so we substitute

(67.21 - L) / L = 398.72 / 110000

110000( 67.21 - L) = 398.72L

7393100 - 110000L = 398.72L

7393100 = 398.72L+ 110000L

7393100 = 110398.72L

L = 7393100 / 110398.72

L = 66.97 mm

Therefore; length of the specimen is 66.97 mm

Answer: Why is even here then.

Explanation:

Answer:

False

Explanation:

I got it wrong picking true

Answer:

Tech A is correct.          

Explanation:

A front-wheel-drive pulling to the left can result from several factors. One of them is definitely a faulty break.

A correct diagnosis linking the problem to the brakes is when there is an internal restriction and the pull is constant to one side and gets worse when the brakes are applied.

To confirm this, one would need to lift the vehicle and rotate each wheel by hand to check for excessive friction.

So the restriction may be caused by:

Cheers