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3 years ago

Can I get some help please? Thanks!

Physics

1 answer:

Brums [2.3K]3 years ago

5 0

To begin with it was believed that light transmitted instantaneously. However now we know that light just moved to fast to be seen.
The first successful measurement was in 1676, by a man named Roemer. He came up with a measurement that light moved a 214,000 km/s. Which was a very approximate evaluation.
Another measurement was obtained in 1728, by a man named James Bradley, he came up with his measurement by observing stellar aberrations. He came up with a value of 301,000 km/s.
In 1849 Fizeau, he used a beam of light and a mirror, he got a value of 315,000 km/s. Foucault improved this measurement a year later, using rotating mirrors he got a more accurate value of 298,000 km/s.

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4 years ago

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3 years ago

The resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is:a) 180 d

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Answer:

0 degrees

Explanation:

Let $F_1\ and\ F_2$ are two forces. The resultant of two forces acting on the same point is given by :

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos\theta}$

Where $\theta$ is the angle between two forces

When $\theta=0$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos(0)}$

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2}$

When $\theta=90^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(90)}$

$F_R=\sqrt{F_1^2+F_2^2}$

When $\theta=180^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(180)}$

$F_R=\sqrt{F_1^2+F_2^2-2F_1F_2}$

It is clear that the resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is 0 degrees. Hence, this is the required solution.

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