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Blizzard [7]
3 years ago
5

Which of the following STEM discoverers is known for inventing the circular saws used in sawmills? Babbitt Edison Gates Hawking

Physics
2 answers:
Jet001 [13]3 years ago
5 0
Among the STEM discoverers, the one who is known for the invention of the circular saws that are used in sawmills is Babbitt or better known as Sarah "Tabitha<span>" Babbitt, an American inventor and Shaker tool maker. The answer to this is the first option. Hope this helps.</span>
lara31 [8.8K]3 years ago
5 0

Answer:

Babbitt

Explanation:

Large circular saws are found in sawmills and are used for cutting wood in most applications. In 1813, Shaker's sister, Tabitha Babbitt invented the first circular saw, noted in a sawmill that was working on the spinning house in the Shaker Harvard community in Massachusetts, when she decided to invent an improvement for two-handed saws. men, for the production of lumber. Tabitha Babbitt is also known as the creator of an improved version of clipped nails, a new method of making false teeth, and a better spinning head.

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A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top
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Answer:

y_{max} = 0.829\,m

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}

v = 2.913\,\frac{m}{s}

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}

After some algebraic handling, a second-order polynomial is formed:

12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s

1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0

The roots of the polynomial are, respectively:

\Delta s_{1} \approx 0.128\,m

\Delta s_{2} \approx -0.099\,m

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

\Delta s \approx 0.128\,m

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