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3 years ago

The visible spectrum refers to the portion of the electromagnetic spectrum that we ________.

Physics

1 answer:

yKpoI14uk [10]3 years ago

8 0

<h2>Answer: can see</h2>

Explanation:

The portion visible by the human eye of the electromagnetic spectrum is between 380 nm (violet-blue) and 780 nm (red) approximately. Which means this part of the spectrum is located between ultraviolet light and infrared light.

Note the fact only part of the whole electromagnetic spectrum is visible to humans is because the receptors in our eyes are only sensitive to these wavelengths.

Therefore:

<h2>The visible spectrum refers to the portion of the electromagnetic spectrum that <u>we </u><u>can see</u></h2>

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Explain the value or importance of using the international system of units when describing motion

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Light that has a wavelength of 554 nm is incident on a single slit that is 10 μm wide. Determine the angular location of the fir

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3 years ago

Suppose that, while lying on a beach near the equator of a far-off planet watching the sun set over a calm ocean, you start a st

zubka84 [21]

Answer:

R=3818Km

Explanation:

Take a look at the picture. Point A is when you start the stopwatch. Then you stand, the planet rotates an angle α and you are standing at point B.

Since you travel 2π radians in 24H, the angle can be calculated as:

$\alpha =\frac{2*\pi *t}{24H}$ t being expressed in hours.

$\alpha =\frac{2*\pi *11.9s*1H/3600s}{24H}=0.000865rad$

From the triangle formed by A,B and the center of the planet, we know that:

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2 years ago

Consider a runner in lane 1 with a speed v1 and a runner in lane 4 with a speed v4 , where the ratio of the speeds v4/v1 is 2 .

kondor19780726 [428]

Answer:

$F_4=\frac{4}{b}F_1$

Explanation:

Centripetal force is the net force acting on a body which makes it move along a curved path. This force is always towards the center of curvature.

The centripetal force is given by:

F = mv² / r

where m is the mass of the body, v is the velocity of the body, r is the radius, F is the centripetal force and v²/r is the centripetal acceleration.

Given that:

$\frac{v_4}{v_1}=2\\\\v_4=2v_1\\\\ Also, radius\ of\ lane\ 4(r_4)=2*radius\ of\ lane\ 1\\\\Let\ r_1\ be\ radius\ of\ lane\ 1,m=mass\ of\ runner\ in\ lane\ 1=mass\ of\ runner\ in\ lane\ 4\ \\and\ v_4=velocity\ of\ runner\ in\ lane\ 4.\ Hence:\\\\r_4=b*r_1=br_1\\\\The\ centripetal\ force\ for\ lane\ 4(F_4)\ is:\\\\F_4=\frac{mv_4^2}{r_4}\\\\ F_4=\frac{m(2v_1)^2}{br_1}\\\\F_4= \frac{4mv_1^2}{br_1} \\\\But\ F_1=\frac{mv_1^2}{r_1}\\\\Hence\ F_4=\frac{4}{b}F_1$

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