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Katen [24]
3 years ago
14

What is the midpoint of a line segment with endpoints at (-3.5, -2.1) and (5.7,3.3)?

Mathematics
1 answer:
Tanzania [10]3 years ago
8 0

Answer:

(1.1 , 0.6)

Step-by-step explanation:

midpoint (x , y): ( x+x') /2 , (y+y') / 2

x = (-3.5 + 5.7) / 2 = 2.2 / 2 = 1.1

y = (-2.1 + 3.3) / 2 = 1.2 / 2 = 0.6

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Three forces act on an object. Two of the forces are at an angle of 95◦ to each other and have magnitudes 35 N and 7 N. The thir
Cerrena [4.2K]

We want to find \vec F_4 such that the object needs is in equilibrium:

\vec F_1+\vec F_2+\vec F_3+\vec F_4=\vec0

We're told that F_1=35\,\mathrm N, F_2=7\,\mathrm N, and F_3=9\,\mathrm N. We also know the angle between \vec F_1 and \vec F_2 is 95º, which means

\vec F_1\cdot\vec F_2=F_1F_2\cos95^\circ=245\cos95^\circ

\vec F_3 is perpendicular to both \vec F_1 and \vec F_2, so \vec F_1\cdot\vec F_3=\vec F_2\cdot\vec F_3=0.

If we take the dot product of \vec F_1 with the sum of all four vectors, we get

\vec F_1\cdot(\vec F_1+\vec F_2+\vec F_3+\vec F_4)=0

\vec F_1\cdot\vec F_1+\vec F_1\cdot\vec F_2+\vec F_1\cdot\vec F_3+\vec F_1\cdot\vec F_4=0

{F_1}^2+\vec F_1\cdot\vec F_2+0+\vec F_1\cdot\vec F_4=0

\implies\vec F_1\cdot\vec F_4=-\left({F_1}^2+\vec F_1\cdot\vec F_2\right)

We can do the same thing with \vec F_2 and \vec F_3:

\vec F_2\cdot(\vec F_1+\vec F_2+\vec F_3+\vec F_4)=0

\implies\vec F_2\cdot\vec F_4=-\left(\vec F_1\cdot\vec F_2+{F_2}^2\right)

\vec F_3\cdot(\vec F_1+\vec F_2+\vec F_3+\vec F_4)=0

\implies\vec F_3\cdot\vec F_4=-{F_3}^2

Finally, if we do this with \vec F_4, we get

\vec F_4\cdot(\vec F_1+\vec F_2+\vec F_3+\vec F_4)=0

\implies{F_4}^2=-\left(\vec F_1\cdot\vec F_4+\vec F_2\cdot\vec F_4+\vec F_3\cdot\vec F_4\right)

\implies{F_4}^2=-\left(-\left({F_1}^2+\vec F_1\cdot\vec F_2\right)-\left(\vec F_1\cdot\vec F_2+{F_2}^2\right)-{F_3}^2\right)

\implies F_4=\sqrt{{F_1}^2+{F_2}^2+{F_3}^2+2(\vec F_1\cdot\vec F_2)}

\implies\boxed{F_4\approx36.2\,\mathrm N}

7 0
3 years ago
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