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3 years ago

Isaac Newton used a ____ to break white light into its component colors.

Physics

2 answers:

ivann1987 [24]3 years ago

8 0

<h2>Answer: Prism</h2>

In the eighteenth century Isaac Newton found out that <u>when a beam of light from the Sun, passes trhough a prism is decomposed in many different colors</u>. He named this phenomenom as dispersion of light.

This phenomenom occurs when a beam of white light (which is compound of many wavelengths or "colors") is refracted (the different rays of light are diverted depending on their wavelengths) in some medium, leaving their constituent colors separated.

Therefore:

<h2>Isaac Newton used a <u>prism</u> to break white light into its component colors.</h2>

Anton [14]3 years ago

7 0

Answer:

Prism APEX Verified

Explanation:

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Explanation: The window is a rectangle with area:

A = 1.1(1.5)

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3 years ago

Calculate the mag-netic field (magnitude and direc-tion) at a point p due to a current i=12.0 a in the wire shown in fig. p28.68

creativ13 [48]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude is $B= 4.2 *10^ {-6}T$ , the direction is into the page

Explanation:

From the question we are told that

The current is $i = 12.0 A$

The radius of arc bc is $r_{bc} = 30.0 \ cm =\frac{30}{100} = 0.3m$

The radius of arc da is $r_{da} = 20.0 \ cm = \frac{20}{100} = 0.20 \ m$

The length of segment cd and ab is = $l = 10cm = \frac{10}{100} = 0.10 m$

The objective of the solution is to obtain the magnetic field

Generally magnetic due to the current flowing in the arc is mathematically represented as

$B = \frac{\mu_o I}{4 \pi r}$

Here I is the current

$\mu_o$ is the permeability of free space with a value of $4\pi *10^{-7}T \cdot m/A$

r is the distance

Considering Arc da

$B_{da} = \frac{\mu_o I}{4 \pi r_{da}} \theta$

Where $\theta$ is the angle the arc da makes with the center from the diagram its value is $\theta = 120^o = 120^o * \frac{\pi}{180} = \frac{2\pi}{3} rad$

Now substituting values into formula for magnetic field for da

$B_{da} = \frac{4\p *10^{-7} * 12}{4 \pi (0.20)}[\frac{2 \pi}{3} ]$

$= \frac{10^{-7} * 12}{0.20} * [\frac{2 \pi}{3} ]$

$B_{da}= 12.56*10^{-6} T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to da is into the pages of the paper

Considering Arc bc

$B_{bc} = \frac{\mu_o I}{4 \pi r_{bc}} \theta$

Substituting value

$B_{bc} = \frac{4 \pi *10^{-7} * 12}{4 \pi (0.30)} [\frac{2 \pi}{3} ]$

$B_{bc}= 8.37*10^{-6}T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to bc is out of the pages of the paper

Since the line joining P to segment bc and da makes angle = 0°

Then the net magnetic field would be

$B = B_{da} - B{bc}$