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EleoNora [17]
3 years ago
10

A team of runners is needed ti run 1/4- mile relay race.if each runner must run 1/16 mile,how many runners will be needed

Mathematics
2 answers:
Pavel [41]3 years ago
4 0

If each runner must run 1/16 of 1/4 mile then you need 4 runners because four runners makes 4/16 points. this number is equivalent to 1/4 which proves this statement true

barxatty [35]3 years ago
3 0
You divide 1/4 by 1/16

KCF- 1/4 x 16/1 = 16/4 = 4

You need 4 runners total, I hope I helped
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I'm using the clip from link [1] as a reference, i.e. the shot of the Tantive IV escaping the Devastator towards the camera, which lasts approximately 5 seconds (2:27 - 2:32). I recorded the particular shot, then carried out a cursory frame-by-frame analysis. The shot itself consists of 371 frames.

At around frame 15 - about 0.233s into the clip - one can barely make out a green projectile being launched from one of the Devastator's batteries. It takes the projectile between 19 and 20 frames (I'll round up to 20) - now at frame 35, about 0.333s later - for it to reach the left edge of the camera's field of view.

Let's suppose the Tantive IV was situated in roughly the same physical position in frame 1 as the projectile was in frame 35.

Now, the Tantive IV exits the screen at around frame 182 - about 3.017s into the clip. According to link [2], the standard CR90 "Corellian" Corvette can reach speeds of up to 950km/h. The Rebels are fleeing the Empire, so it's reasonable to assume that they are moving as fast as they can. Let's also assume they're traveling at a constant rate, and that the tractor beam aboard the Devastator hasn't been activated yet.

Note: Link [2] refers to this maximum speed as "atmosphere speed". I can't find any references as to what this means, so I assume it means something along the lines of "same conditions as Earth's atmosphere". The two ships are flying above Tatooine, which is inhabited in part by humans, so it's probably a safe bet that the atmospheres of Tatooine and Earth aren't that dissimilar.

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2.296\text{ km}=r\times(0.333\text{ s})\implies r\approx24,821.6\dfrac{\text{km}}{\text{h}}
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