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1 year ago

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An intake manifold gasket has been replaced due to a vacuum leak. Which of the following steps uses a scan tool to complete the

job? O A. Torquing the manifold bolts B. Idle relearn O C. Refilling the cooling system O D. Air cleaner check

1 answer:

matrenka [14]1 year ago

7 0

Answer: B.Idle relearn

Explanation:

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An aluminum bar 125 mm long with a square cross section 16 mm on an edge is pulled in tension with a load of 66,700 N and experi

AfilCa [17]

Answer: the modulus of elasticity of the aluminum is 75740.37 MPa

Explanation:

Given that;

Length of Aluminum bar L = 125 mm

square cross section s = 16 mm

so area of cross section of the aluminum bar is;

A = s² = 16² = 256 mm²

Tensile load acting the bar p = 66,700 N

elongation produced Δ = 0.43

so

Δ = PL / AE

we substitute

0.43 = (66,700 × 125) / (256 × E)

0.43(256 × E) = (66,700 × 125)

110.08E = 8337500

E = 8337500 / 110.08

E = 75740.37 MPa

Therefore, the modulus of elasticity of the aluminum is 75740.37 MPa

4 0

2 years ago

Water is pumped from a lake to a storage tank 18 m above at a rate of 70 L/s while consuming 20.4 kW of electric power. Disregar

suter [353]

Search up A gardener can increase the number of dahlia plants in an annual garden by either buying new bulbs each year or dividing the existing bulbs to create new plants . The table below shows the expected number of bulbs for each method

Part A
For each method,a function to model the expected number of plants for each year

Part B
Use the Functions to Find the expected number of plants in 10 years for each method.

Part C

3 0

2 years ago

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5×10-4

krok68 [10]

Answer:

1788.9 MPa

Explanation:

The magnitude of the maximum stress (σ) can be calculated usign the following equation:

$\sigma = 2\sigma_{0} \sqrt{\frac{a}{\rho}}$

<u>Where:</u>

<em>ρ: is the radius of curvature = 2.5x10⁻⁴ mm (0.9843x10⁻⁵ in)</em>

<em>σ₀: is the tensile stress = 100x10⁶ Pa (14500 psi) </em>

<em>2a: is the crack length = 4x10⁻² mm (1.575x10⁻³ in) </em>

Hence, the maximum stress (σ) is:

$\sigma = 2*100\cdot 10^{6} Pa \sqrt{\frac{(4 \cdot 10^{-2} mm)/2}{2.5 \cdot 10^{-4} mm}} = 1.79 \cdot 10^{6} Pa = 1788.9 MPa$

Therefore, the magnitude of the maximum stress is 1788.9 MPa.

I hope it helps you!

5 0

3 years ago

Using Von Karman momentum integral equation, find the boundary layer thickness, the displacement thickness, the momentum thickne

Alex_Xolod [135]

Answer:

Explanation:

We can solve Von Karman momentum integral equation as seen below using following in the attached file

3 0

2 years ago

In a production turning operation, the foreman has decreed that a single pass must be completed on the cylindrical workpiece in

stellarik [79]

Answer:

V = 125.7m/min

Explanation:

Given:

L = 400 mm ≈ 0.4m

D = 150 mm ≈ 0.15m

T = 5 minutes

F = 0.30mm ≈ 0.0003m

To calculate the cutting speed, let's use the formula :

$T = \frac{pi* D * L}{V*F}$

We are to find the speed, V. Let's make it the subject.

$V = \frac{pi* D * L}{F*T}$

Substituting values we have:

$V = \frac{pi* 0.4 * 0.15}{0.0003*5}$

V = 125.68 m/min ≈ 125.7 m/min

Therefore, V = 125.7m/min

7 0

2 years ago