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MrRa [10]
3 years ago
5

The development of various technologies led to many historic events. Use information from the Internet to describe one major his

toric event
initiated by a technological development or discovery

-


Sometimes, world events lead to the development of a technology. Perform online research and identify a historic event that required the

development of technologies. Discuss one technology created because of this event. How does it continue to affect human lives today?
Engineering
1 answer:
11111nata11111 [884]3 years ago
4 0

Answer:

1. Industrial revolution was initiated or borne through the production of Steel

2. World War 1 led to the development of Tanks

Explanation:

The production of Steel through the Bessemer Process in the middle of the nineteenth century was a major technological development that spurred the Industrial revolution. This invention led to the widespread use of steel in the production of many things including vehicles and airplanes.

During the First World War in 1914, soldiers found the use of just their armaments in battle as not so productive. This led to the development of Tanks in 1915 that would continue moving towards the enemy even when being shot at.

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What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9 × 10
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Answer:

Recall the formula for the maximum stress, σₐ = 2σ₀ *√ (α/ρₓ)

where

σ₀ = tensile stress = 140 MPa = 1.40x 10⁸Pa

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Substituting these values into the formula, we can calculate the max stress as

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3 years ago
**Please Help, ASAP**
Novay_Z [31]

Explanation:

We need to rearrange the following formula for the values given in parenthesis.

(1) x+xy = y, (x)

taking x common in LHS,

x(1+y)=y

x=\dfrac{y}{1+y}

(2) x+y = xy, (x)

Subrtacting both sides by xy.

x+y-xy = xy-xy

x+y-xy = 0

x-xy=-y

x(1-y)=-y

x=\dfrac{-y}{1-y}

(3) x = y+xy, (x)

Subrating both sides by xy

x-xy = y+xy-xy

x(1-y)=y

x=\dfrac{y}{1-y}

(4) E = (1/2)mv^2-(1/2)mu^2, (u)

Subtracting both sides by (1/2)mv^2

E-(1/2)mv^2 = (1/2)mv^2-(1/2)mu^2-(1/2)mv^2

E-(1/2)mv^2 =-(1/2)mu^2

So,

2(E-\dfrac{1}{2}mv^2)=-mu^2\\\\u^2=\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)\\\\u=\sqrt{\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)}\\\\u=\sqrt{\dfrac{2}{m}(\dfrac{1}{2}mv^2-E)}

(5) (x^2/a^2)-(y^2/b^2) = 1, (y)

\dfrac{x^2}{a^2}-1=\dfrac{y^2}{b^2}\\\\y^2=b^2(\dfrac{x^2}{a^2}-1)\\\\y=b\sqrt{\dfrac{x^2}{a^2}-1}

(6) ay^2 = x^3, (y)

y^2=\dfrac{x^3}{a}\\\\y=\sqrt{\dfrac{x^3}{a}}

Hence, this is the required solution.

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Explanation:

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