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4 years ago

What did the Spanish have that the Aztecs didn’t

Engineering

1 answer:

notsponge [240]4 years ago

6 0

Answer:

The Spanish brought chickenpox, smallpox, measles, mumps, and rubella to the new world. These diseases killed 75% of the Native Americans. ... The advantages that the Spanish had over the Aztecs were 16 horses, guns, armor, formed alliances, and diseases, steel.

Explanation:

I hope this helped!

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Write down one metal or alloy that is best suited for each of the following applications:

777dan777 [17]

Answer:

Explanation:

a. Cast iron or Aluminium alloy are typically used. Aluminium is much lighter in weight and it can transfer heat better to the coolant. While Cast Iron is typically stronger and is thus still used by the manufacturers.

b. Copper can be used as a condensing heat exchanger for hot steam due to its optimal thermal properties and its ability to resist corrosion.

c. high-speed steel are perfect for producing drill bits because of its hardness and resistance to heat to an extent. Drill bits tend to produce heat as a result of the friction between it and the material to be drilled.

d. lead can be used as a container for strong acids because of its anti-corrosive properties

e.zinc and copper can be used as fuel in pyrotechnics mainly due to the fact that burn with refreshing colours. Aluminium can also be used.

f. Platinum is the metal that best suits this purpose because of its high melting point and resistivity to oxidation.

6 0

3 years ago

A circular section of material is tested. The original specimen is 200 mm long and has a diameter of 13 mm. When loaded to its p

n200080 [17]

Answer:

modulus of elasticity = 100.45 Gpa,

proportional limit = 150.68 N/mm^2.

Explanation:

We are given the following parameters or data in the question as;

=> "The original specimen = 200 mm long and has a diameter of 13 mm."

=> "When loaded to its proportional limit, the specimen elongates by 0.3 mm."

=> " The total axial load is 20 kN"

Step one: Calculate the area

Area = π/ 4 × c^2.

Area = π/ 4 × 13^2 = 132.73 mm^2.

Step two: determine the stress induced.

stress induced = load/ area= 20 × 1000/132.73 = 150.68 N/mm^2.

Step three: determine the strain rate:

The strain rate = change in length/original length = 0.3/ 200 = 0.0015.

Step four: determine the modulus of elasticity.

modulus of elasticity = stress/strain = 150.68/0.0015 = 100453.33 N/mm^2 = 100.45 Gpa.

Step five: determine the proportional limit.

proportional limit = 20 × 1000/132.73 = 150.68 N/mm^2.

7 0

3 years ago

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What is a problem that technology can help solve that problem?

Maslowich

Seeing what the other side of the world is doing right now

8 0

3 years ago

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Các đặc điểm chính của đường dây dài siêu cao áp .

rodikova [14]

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

8 0

3 years ago

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

4 0

4 years ago