What did Congress do in 1787 to settle land disputes among the settlers?

Answer:

A turbo blade has the best features from both other types of blade. The continuous, serrated edge makes for fast cutting while remaining smooth and clean. They are mostly used to cut a variety of materials, such as tile, stone, marble, granite, masonry, and many other building materials.

Explanation:

Answer:

Masonry Saw

Explanation:

Masonry Saws can be used to cut brick, ceramic, tile and or stone.

Answer:

false

Explanation:

Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as cold forging.​

Forging is a metalworking process that involves shaping metal with localized compressive forces. A hammer or a die is used to deliver the blows.

Forging is frequently classified according to temperature: cold forging, warm forging, or hot forging.

The goal of forging is to make metal parts. Metal forging produces some of the most durable manufactured parts available when compared to other manufacturing methods.

Minor cracks and empty spaces in the metal are filled as the metal is heated and pressed.

Cold forging is the process of deforming a metal material at room temperature using extremely high pressure. The slug is placed in a die and compressed by a press until it fits into the desired shape.

Thus, the answer is cold forging.

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Answer:

B a cast part

Explanation:

Extrusion is defined as the process of shaping material, such as aluminum, by forcing it to flow through a shaped opening in a die. Extruded material emerges as an elongated piece with the same profile as the die opening.

Forcing a solid piece of heated aluminum through a die forms: a cast part. Hence, option B is correct.

A liquid element is more often filled with concrete that has a hollow chamber in the correct form during the casting manufacturing process, and the item is then let to harden.

A casting, which is the term for the solidified component, is ejected or broken out of the mould to complete the procedure.

Thus, option B is correct.

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Answer:

The annualized cost is:

$299,272.

Explanation:

a) Data and Calculations:

Year 0  Capital cost of road construction = $6 million

Year 10 Major improvements cost = $17 million

Year 25 Replacement and upgrade cost = $29 million

Year 40 Federal government one-time tax credit = $12 million

Period of project analysis = 50 years

Cost of capital (discount rate) = 10%

Annualized cost at present value costs:

Amount spent  Discount Factor     Present value

$6 million           1                               $6,000,000

$17 million         0.386                          6,562,000

$29 million       0.092                          2,668,000

($12 million)      0.0222                         (266,400)

Total cost                                          $14,963,600

Annualized cost = $14,963,600/50 =  $299,272

b) The annualized cost for the road construction project, which is the annualized value of the net present costs of $14,963,600, is divided by 50.  Before obtaining the net present costs, the cash outflows, including the tax credit, are discounted to their present values, using the discount rate of 10%.  And then, the average of the cost is obtained by dividing the total net present cost into 50 years.

Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, [tex]\tau_{max}[/tex], is given by the following formula;

[tex]\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )[/tex]

[tex]t_w[/tex] = 1 cm = 0.01

h = 29 cm = 0.29 m

[tex]h_w[/tex] = 25 cm = 0.25 m

b = 15 cm = 0.15 m

[tex]I_c[/tex] = The centroidal moment of inertia

[tex]I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )[/tex]

[tex]I_c[/tex] = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

[tex]I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right ) = 1.2257083\bar 3 \times 10^{-4}[/tex]

[tex]I_c[/tex] = 1.2257083[tex]\bar 3[/tex] × 10⁻⁴ m⁴

From which we have;

[tex]4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )[/tex]

Which gives;

W = 11,416.6879 N

[tex]\sigma _{b.max} = \dfrac{M_c}{I_c}[/tex]

[tex]\sigma _{b.max}[/tex] = 1500 N/cm² = 15,000,000 N/m²

[tex]M_c[/tex] = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

[tex]M_{max} = \dfrac{W \cdot L}{4}[/tex]

[tex]L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417[/tex]

L ≈ 0.64417 m ≈ 64.417 cm.

Answer:

a) for back pressures of (a) 500 kPa

- mass flow rate is 2.127 kg/s

- exit Mach number is 1.046

b) for back pressures of (a) 784 kPa

- mass flow rate is 1.793 kg/s

- exit Mach number is 0.6

Explanation:

Given that;

A₂ = 0.001 m²

P₁ = 1 MPa

T₁ = 360 K

k = 1.4

P₂ = 500 Kpa

(1000/500)^(1.4-1 / 1.4) = 360 /T₂

2^(0.4/1.4) = 360/T₂

1.219 = 360 / T₂

T₂ = 360 / 1.219

T₂ = 295.32 K

CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000

we substitute

CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000

1.005 × 360 =  1.005 × 295.32 + v₂²/2000

v₂ = 360.56 m/s²

p₂v₂ = mRT₂

500 × (0.001 × 360.56) = m × 0.287 × 295.32

m = 2.127 kg/s

so Mach Number = V₂ / Vc

Vc = √( kRT) = √( 1.4 × 287 × 295.32) = 344.47 m/s

So Mach Number =  V₂ / Vc  =  360.56 / 344.47 = 1.046

Therefore for back pressures of (a) 500 kPa

- mass flow rate is 2.127 kg/s

- exit Mach number is 1.046

b)

AT P₂ = 784 kPa

(1000/784)^(1.4-1 / 1.4) = 360/T₂

T₂ = 335.82 K

now

V₂²/2000 = 1.005( 360 - 335.82)

V₂ = 220.45 m/s

P₂V₂ = mRT₂

784 × (0.001 × 220.45) = m( 0.287) ( 335.82)

172.83 = 96.38 m

m = 172.83 / 96.38

m = 1.793 kg/s

just like in a)

Vc = √( kRT) = √( 1.4 × 287 × 335.82) = 367.32 m/s

Mach Number = V₂ / Vc = 220.45 / 367.32 = 0.6

Therefore for back pressures of (a) 784 kPa

- mass flow rate is 1.793 kg/s

- exit Mach number is 0.6

Following are the

Given:

[tex]A_2 = 0.001\ m^2\\\\P_1 = 1\ MPa\\\\T_1 = 360\ K\\\\k = 1.4\\\\P_2 = 500\ Kpa\\\\[/tex]

To find:

Flow rate of mass, and Mach number

Solution:

For point a)

Using formula:

[tex]\to P^{\frac{r-1}{r}} \alpha T\\\\\to (\frac{1000}{500})^(\frac{1.4-1}{1.4}) = \frac{360}{T_2}\\\\\to 2^(\frac{0.4}{1.4}) = \frac{360}{T_2}\\\\\to 1.219 = \frac{360}{T_2}\\\\\to T_2 = \frac{360}{1.219}\\\\\to T_2 = 295.32\ K\\\\[/tex]

[tex]\to C_pT_1 + \frac{V_1^2}{2000} = C_pT_2 + \frac{V_2^2}{2000}\\\\\to 1.005 \times 360 = 1.005 \times 295.32 + \frac{v_2^2}{2000}\\\\\to v_2 = 360.56 \ \frac{m}{s^2} \\\\\to p_2v_2 = mRT_2\\\\\to 500 \times (0.001 \times 360.56) = m \times 0.287 \times 295.32\\\\\to m = 2.127\ \frac{kg}{s}\\\\[/tex]

Mach Number [tex]= \frac{V_2}{V_c}\\\\[/tex]

[tex]\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 295.32)} = 344.47 \ \frac{m}{s}\\\\[/tex]

[tex]\to \frac{V_2}{V_c} = \frac{360.56}{344.47} = 1.046[/tex]

For point b)

[tex]\to P_2 = 784\ kPa\\\\\to (\frac{1000}{784})^{(\frac{0.4}{1.4})} = \frac{360}{T_2}\\\\\to T_2 = 335.82\ K\\\\[/tex]

now

[tex]\to \frac{V_2^2}{2000} = 1.005( 360 - 335.82)\\\\\to V_2 = 220.45 \frac{m}{s}\\\\\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 335.82)} = 367.32\ \frac{ m}{s}\\\\\to c= \frac{220.45}{ 367.32} = 0.6\\[/tex]

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