@cvbearden21 asked 6 months ago.

How much sooner does the box reach the bottom of the incline than the disk?

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@logic answered 5 months ago.

 

now s is the length of the incline in terms of given variables, s = h/sin(theta) time to bottom, box, t_box = 2h/(sin(theta)*sqrt(2gh)) time to bottom, box, t_disk = 2h/(sin(theta)*sqrt(4gh/3))

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@Gaurii answered 6 months ago.

now s is the length of the incline in terms of given variables, s = h/sin(theta) time to bottom, box, t_box = 2h/(sin(theta)*sqrt(2gh)) time to bottom, box, t_disk = 2h/(sin(theta)*sqrt(4gh/3))

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@clarkee answered 6 months ago.

 a particular type of cell doubles in number every hour. which function can be used to find the number of cells present at the end of h hours if there are initially 4 of these cells? a. n = 4 ()" b. n = 4(2)" c. n = 4 + (2)" din = 4 +"   Answers: 1
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@omeppereku answered 6 months ago.

First, let’s take a look at the image.  The box falls a greater distance than the disk because it’s much heavier.  The box has a mass of 1,000g, while the disk has a mass of 10g.
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@eddenubiss answered 6 months ago.

First, let's take a look at the diagram below (Figure 1). The incline and the disk are at the same point when the disk is at the highest point on the incline.
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