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3 years ago

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What did isaac Newton believe light was made of

2 answers:

adelina 88 [10]3 years ago

8 0

Sir Isaac Newton, held the theory that light was made up of tiny particles<span>. In 1678, Dutch physicist, Christiaan Huygens, believed that light was made up of </span>waves<span>vibrating up and down </span>perpendicular<span> to the direction of the light travels, and therefore formulated a way of visualising wave propagation.</span>

STALIN [3.7K]3 years ago

5 0

Answer:

PARTICLE

Explanation:

As per the theory proposed by the Newton for the propagation of light he assumed that light is made up of PARTICLEs

He said that light is made up of small rigid particles which behaves elastically. When light shows reflection then this is an example of 100% elastic collision where these particles will strike the surface and rebound with same speed at same angle with the normal

So this theory explains the propagation of light in particle form and hence this theory was known as particle theory of light

So as per Newton's explanation light is made up of PARTICLES

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Explanation:

It is given that,

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The component along her direction of motion is shown in attached figure. It means

$W_y=mg\sin\theta\\\\W_y=55\times 9.8\times \sin(32)\\\\W_y=285.62\ N$

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Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

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