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-Dominant- [34]
3 years ago
5

For trapezoid JKLM, A and B are midpoints of the legs. Find AB.

Physics
1 answer:
kodGreya [7K]3 years ago
7 0

Answer:

AB = 29

Explanation:

For a better understanding, we must work this problem in a graphic way. In the attached image we can see the solution.

First, we draw a vertical dotted lines from the point J & K to the line ML, then we can see two new portions with the same length. Then with this simple analysis:

2x = 39 - 19

x = 10

Then we know that x = 10, another important data to find the answer is that the AB line is located in the midpoints of the legs. We also can see the right triangle MJ and the dotted line.

Now for every single right triangle, no matter its size and relationship between the vertical and the horizontal lengths, if some point is located in the hypotenuse (leg) at the middle of its length. This will be proportional to the vertical and the horizontal cathetus, therefore we will have the middle point on those two lines.

So, the AB line will be the sum of JK plus two times 5

AB = 19 + 5 + 5 = 29

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Answer:

a= 3.49 m/s^2

Explanation:

magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.

we know that tangential acceleration a_t= change in velocity /time taken

now 90 km/h = 25 m/s

a_t = 25/17 = 1.47 m/s^2.

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v= a_t×t = 1.47×13 = 19.11 m/s

a_r = 19.11^2/115=  3.175

now,

a= \sqrt{a_t^2+a_r^2}

a= \sqrt{1.47^2+3.175^2}

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3 0
3 years ago
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Answer:

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Answer: 3.906kg

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