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goldenfox [79]
3 years ago
15

What major advancement in machine tools occurred in the 1970s and what benefits did it provide? describe in your own words.

Engineering
2 answers:
mixer [17]3 years ago
4 0

Answer:

I'm just a seventh grader

kolbaska11 [484]3 years ago
4 0

Answer:

here's an essay I wrote one this topic

Explanation:

The early 1970s saw price and wage freezes and high unemployment. The oil crisis of 1973 and 1979 served to help slow an already snail-paced economy, and by 1979, the U.S. was experiencing the worst inflation in 33 years.

A growing world market was not helping the economic transition as traditionally strong big businesses like automotive and steel lost market shares to Europe and Japan. At the beginning of the decade, the government bailed out Lockheed Martin with $250 million worth of loan guarantees. Later it did the same for Chrysler—to the tune of $1 billion.

Obviously, the old ways of doing business were failing. The seeds of concern about more efficient operations, more flexibility, and higher quality were taking root in this decade for the eventual growth of a new paradigm in manufacturing that would flourish in the 80s. As NCs and DNCs gained recognition and popularity, the benefits of faster, more accurate cutting began to make demands on traditional tooling, controls, inspection, and other facets of the manufacturing process. This led to the development of many new companies who were ready to meet the challenges of a transitioning decade and make the products that would fulfill the needs of a new world of manufacturing.

As manufacturers looked for inexpensive solutions to their problems, those in the machine tool industry responded with new, lower-cost technologies. GE Fanuc Automation for one developed its Mark Century 550 Series in 1970. The highly-reliable control system was marketed as a complex, low-cost NC and became popular in the worldwide market. It introduced new features like lathe canned cycles, spindle speed monitor, battery backed tool offset, RAM memory, and cutter compensation.

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About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa
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The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

K.E = \frac{2KZe^2}{r}

where;

  • Z is the atomic number of aluminium  = 13
  • e is charge
  • r is distance of closest approach = thickness of aluminium
  • k is Coulomb's constant = 9 x 10⁹ Nm²/C²
<h3>For 2.5 MeV electrons</h3>

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

Learn more about closest distance of approach here: brainly.com/question/6426420

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