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3 years ago

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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Mathematically, F= ma where m is mass and a is acceleration

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2 years ago

A train traveled 85 mph for 7 hours how far did it travel?

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2 years ago

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A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

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$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

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$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

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$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

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2 years ago

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nirvana33 [79]

It gets larger because
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3 years ago

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