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patriot
7 days ago
15

How did Newton use creativity and logic in his approach to investigating light?

Physics
2 answers:
Softa [2K]7 days ago
8 0
Isaac Newton displayed creativity by employing prisms to demonstrate that white light comprises various colors. He utilized logic in the rhetorical structure of his arguments to persuade his audience of his conclusions' validity.

He was not the initial experimenter with prisms; French philosopher Rene Descartes had conducted similar experiments. However, Descartes believed that the prism altered the light to create the color spectrum. Newton, on the other hand, accurately recognized that light refracting through a prism exposes the inherent colors within it. He utilized a second prism, filtering out all but one color, to illustrate that a lone color did not change when passing through a prism. By adjusting this second prism, he also proved that the various colors of light could be blended back into white light. Newton's 1672 paper on light refraction through prisms solidified his standing as a scientist, and he continued to explore light throughout his career, publishing a more extensive work in 1704 on Opticks (as optics was spelled at the time).
Sav [2.2K]7 days ago
3 0
He approached the concept of light using quantum mechanics to explain relativity and perspective.
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28 days ago
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21 day ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
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1) 9.18 s

During the initial phase, the rocket accelerates at

a_1=13.5 m/s^2

over a time interval of

t_1=3.50 s

The final velocity after this period is found using the SUVAT formula:

v_1=u+a_1t_1

with initial velocity u = 0. Substituting a1 and t1 gives:

v_1=(13.5)(3.50)=47.3 m/s

Afterward, the rocket decelerates uniformly at

a_2 = -5.15 m/s^2

until it stops, meaning the final velocity is

v_2 = 0

Again using the SUVAT formula,

v_2 = v_1 + a_2 t_2

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2) 299.9 m

Calculate distances covered in both phases.

Distance in the first phase:

d_1 = ut_1 + \frac{1}{2}a_1 t_1^2

Substituting values from part 1,

d_1 = 0 + \frac{1}{2}(13.5) (3.50)^2=82.7 m

Distance in the deceleration phase:

d_2= v_1 t_2 + \frac{1}{2}a_2 t_2^2

Substituting known values,

d_2 = (47.3)(9.18) + \frac{1}{2}(-5.15) (9.18)^2=217.2 m

Total distance traveled is

d = 82.7 m + 217.2 m = 299.9 m

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