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Reika [66]
2 years ago
5

What is valve overlap?

Engineering
1 answer:
Thepotemich [5.8K]2 years ago
6 0

Answer:

Effects of Valve Overlap. As discussed previously, valve overlap is the time when both intake and exhaust valves are open. In simple terms, this provides an opportunity for the exhaust gas flow and intake flow to influence each other.

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2.18 The net potential energy between two adjacent ions, EN, may be represented by the following equation: (1) Calculate the bon
goblinko [34]

Answer:

2.18

Explanation:

because ya correc

3 0
2 years ago
Could I please get help with this​
alex41 [277]

Answer:

1.I_{xc} = 7.161458\overline 3 in.⁴

I_{yc} = 36.661458\overline 3 in.⁴

Iₓ = 28.6458\overline 3 in.⁴

I_y = 138.6548\overline 3 in.⁴

2. I_{xc} = 114.\overline 3 in.⁴

I_{yc} = 37.\overline 3 in.⁴

Iₓ = 457.\overline 3 in.⁴

I_y = 149.\overline 3 in.⁴

3. The maximum deflection of the beam is 2.55552 inches

Explanation:

1. The height of the beam having a rectangular cross section is h = 2.5 in.

The breadth of the beam, is = 5.5 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

I_{xc} = b·h³/12 = 5.5 × 2.5³/12 = 1375/192 = 7.161458\overline 3

I_{xc} = 7.161458\overline 3 in.⁴

I_{yc} = h·b³/12 = 2.5 × 5.5³/12 = 6655/192 = 36.661458\overline 3

I_{yc} = 36.661458\overline 3 in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 5.5 × 2.5³/3 = 625/24 = 28.6458\overline 3

Iₓ = 28.6458\overline 3 in.⁴

I_y = h·b³/3 = 2.5 × 5.5³/3 = 6655/48= 138.6548\overline 3

I_y = 138.6548\overline 3 in.⁴

2. The height of the beam having a rectangular cross section is h = 7 in.

The breadth of the beam, b = 4 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

I_{xc} = b·h³/12 = 4 × 7³/12 = 114.\overline 3

I_{xc} = 114.\overline 3 in.⁴

I_{yc} = h·b³/12 = 7 × 4³/12 = 37.\overline 3

I_{yc} = 37.\overline 3 in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 4 × 7³/3 = 457.\overline 3

Iₓ = 457.\overline 3 in.⁴

I_y = h·b³/3 = 2.5 × 5.5³/3 = 149.\overline 3

I_y = 149.\overline 3 in.⁴

3. The deflection, \delta _{max}, of a simply supported beam having a point load at the center is given as follows;

\delta_{max} = \dfrac{W \times L^3}{48 \times E \times I}

The given parameters of the beam are;

The length of the beam, L = 22 ft. = 264 in.

The applied load at the center, W = 750 lbs

The modulus of elasticity for Cedar = 10,000,000 psi

The height of the wood, h = 3 in.

The breadth of the wood, b = 5 in.

The moment of inertia of the wood, I_{xc} = b·h³/12 = 5 × 3³/12 = 11.25 in.⁴

By plugging in the given values, we have;

\delta_{max} = \dfrac{750 \times 264^3}{48 \times 10,000,000 \times 11.25} = 2.55552

The maximum deflection of the beam, \delta _{max} = 2.55552 inches

5 0
3 years ago
Waste that is generated by a business is called a _____________.
Vinil7 [7]

Answer:

A waste stream

Explanation:

7 0
3 years ago
Read 2 more answers
What is the major drawback to use whiskers as a dispersed agents in composites? a)- High price b)- Large length to diameter rati
artcher [175]

Answer: C) a&c

Explanation: Composite materials are the materials which are made up of the two or more material of different properties.They are usually having properties like high hardness,low in density,no dissolving into each other etc. Whiskers as a dispersing agent in composite materials are usually not preferred because they are expensive as well as they are not easy to disperse in the composite material.Thus option (c) is the correct answer.

7 0
3 years ago
A sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It
topjm [15]

Answer:

a) the moisture content before it was placed in the oven is 18.18%

b) degree of saturation for soil is 72.19%

Explanation:

Given the data in the question;

Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100

so we substitute

Moisture content = [(53.3 - 45.1) / 45.1 ] × 100

= (8.2/45.1) × 100

= 18.18%

Therefore the moisture content before it was placed in the oven is 18.18%

Dry Unit Weight = dry weight / volume

Dry Unit Weight = 45.1 lb / 0.45 ft³

Dry Unit Weight = 100.22 lb/ft³

we know that;

dry unit weight = (Specific gravity × unit weight of water) / (1 + e)

we also know that; unit weight of water is 62.43 lbf/ft³

so we substitute

e = (2.70×62.43 / 100.22) - 1

e = 1.68 - 1

e = 0.68

so void ratio e = 0.68

Now we determine the degree of saturation using the equation;

degree of saturation = (Moisture content × specific gravity) / void ratio

we substitute

degree of saturation = ( 18.18% × 2.7) / 0.68

= 0.49086 / 0.68

= 0.7219 ≈ 72.19%

Therefore degree of saturation for soil is 72.19%

7 0
3 years ago
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