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2 years ago

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

2 answers:

soldier1979 [14.2K]2 years ago

8 0

Isaac Newton was creative in his use of prisms to show how white light is actually made up of multiple colors. He used logic in the way he presented his arguments rhetorically in order to convince readers of the correctness of his conclusions.

Newton was not the first to experiment with passing light through prisms to determine how light works. French philosopher Rene Descartes had done prism experiments of his own. But Descartes had thought that passing through a prism actually modified the light in order to produce the color spectrum. Newton correctly understood that when light refracted through the prism, it revealed the range of colors that were naturally in the light. He then used a second prism, blocking all but one color, to show that a single color passing through a prism was not modified in color. He also showed--by positioning the second prism differently--how the multiple colors of light could be recombined into white light again.

Newton's 1672 paper on light refracting through prisms established his reputation as a scientist. He continued to study light throughout his scientific career, publishing a larger work in 1704 on <em>Opticks</em> (as they spelled optics then).

ryzh [129]2 years ago

3 0

He used the quantum mechanic theory to approach light by describing relativity and perspective.

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A steel cable 1.25 in. in diameter and 50 ft long is to lift a 20-ton load without permanently deforming. What is the length of

Over [174]

Answer:

50.0543248872 ft

Explanation:

F = Load = 20 ton = $20\times 2000\ lb$

d = Diameter = 1.25 in

$L_1$ = Initial length = 50 ft

$L_2$ = Final length

A = Area = $\dfrac{\pi}{4}d^2$

Y = Young's modulus = $30\times 10^6\ psi$

Young's modulus is given by

$Y=\dfrac{FL}{A\Delta L}\\\Rightarrow Y=\dfrac{FL_1}{\dfrac{\pi}{4}d^2(L_2-L_1)}\\\Rightarrow L_2=\dfrac{4FL_1}{Y\pi d^2}+L_1\\\Rightarrow L_2=\dfrac{4\times 40000\times 50}{30\times 10^6\times \pi\times 1.25^2}+50\\\Rightarrow L_2=50.0543248872\ ft$

The length during the lift is 50.0543248872 ft

6 0

2 years ago

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

bixtya [17]

Answer:

1340.2MW

Explanation:

Hi!

To solve this problem follow the steps below!

1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid

W=αhQ

α=specific weight for water =9.81KN/m^3

h=height=220m

Q=flow=690m^3/s

W=(690)(220)(9.81)=1489158Kw=1489.16MW

2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator

Wr=(0.9)(1489.16MW)=1340.2MW

the maximum possible electric power output is 1340.2MW

3 0

1 year ago

Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele

Julli [10]

Answer:

0.018 J

Explanation:

The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

$W = q \Delta V$

where

$q=3.0 \mu C = 3.0 \cdot 10^{-6} C$ is the magnitude of the charge

$\Delta V = 6.0 kV = 6000 V$ is the potential difference between point P and infinity

Substituting into the equation, we find

$W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J$

4 0

2 years ago

Given three capacitors, c1 = 2.0 μf, c2 = 1.5 μf, and c3 = 3.0 μf, what arrangement of parallel and series connections with a 12

Lesechka [4]

Answer:

Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.

$-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-$ .

Explanation:

Consider four possible cases.

$-\begin{array}{c}-{\bf 2.0\;\mu\text{F}-}\\-1.5\;\mu\text{F}- \\-3.0\;\mu\text{F}-\end{array}-$

In case all three capacitors are connected in parallel, the $2.0\;\mu\text{F}$ capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.

$-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-$

In case the $2.0\;\mu\text{F}$ capacitor is connected in parallel with the $1.5\;\mu\text{F}$ capacitor, and the two capacitors in parallel is connected to the $3.0\;\mu\text{F}$ capacitor in series.

The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.

The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,

$\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_3}+ \dfrac{1}{C_1+C_2}} = \frac{1}{\dfrac{1}{3.0}+\dfrac{1}{2.0+1.5}} = 1.62\;\mu\text{F}$ .

What will be the voltage across the 2.0 μF capacitor?

The charge stored in two capacitors in series is the same as the charge in each capacitor.

$Q = C(\text{Effective}) \cdot V = 1.62\;\mu\text{F}\times 12\;\text{V} = 19.4\;\mu\text{C}$ .

Voltage is the same across two capacitors in parallel.As a result,

$\displaystyle V_1 = V_2 = \frac{Q}{C_1+C_2} = \frac{19.4\;\mu\text{C}}{3.5\;\mu\text{F}} = 5.54\;\text{V}$ .

$-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-$ .

Similarly,

the effective capacitance of the two capacitors in parallel is 5.0 μF;
the effective capacitance of the three capacitors, combined: $\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_2}+ \dfrac{1}{C_1+C_3}} = \frac{1}{\dfrac{1}{1.5}+\dfrac{1}{2.0+3.0}} = 1.15\;\mu\text{F}$ .

Charge stored:

$Q = C(\text{Effective}) \cdot V = 1.15\;\mu\text{F}\times 12\;\text{V} = 13.8\;\mu\text{C}$ .

Voltage:

$\displaystyle V_1 = V_3 = \frac{Q}{C_1+C_3} = \frac{13.8\;\mu\text{C}}{5.0\;\mu\text{F}} = 2.76\;\text{V}$ .

$-2.0\;\mu\text{F}-1.5\;\mu\text{F}-3.0\;\mu\text{F}-$ .

Connect all three capacitors in series.

$\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_1} + \dfrac{1}{C_2}+\dfrac{1}{C_3}} =\frac{1}{\dfrac{1}{2.0} + \dfrac{1}{1.5}+\dfrac{1}{3.0}} =0.667\;\mu\text{F}$ .

For each of the three capacitors:

$Q = C(\text{Effective})\cdot V = 0.667\;\mu\text{F} \times 12\;\text{V} = 8.00\;\mu\text{C}$ .

For the $2.0\;\mu\text{F}$ capacitor:

$\displaystyle V_1=\frac{Q}{C_1} = \frac{8.00\;\mu\text{C}}{2.0\;\mu\text{F}} = 4.0\;\text{V}$ .

6 0

1 year ago

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅ = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0

1 year ago