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Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$

So, the y-component of the resultant force is

$F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N$

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

$F=\sqrt{F_x^2+F_y^2}$

Where in this problem, we have:

$F_x=-7.14 N$ is the x-component

$F_y=2.70 N$ is the y-component

And substituting, we find:

$F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N$

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The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

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Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

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An object that looks white when exposed to sunlight reflects all colors of light. What

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It looks blue as it is only reflecting blue light

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suppose you needed to build a raft to cross a fast-moving river. Describe the physical and chemical properties of the raft that

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1.Density of the material building the raft must be lower than the water.
2. Material must not react with water. 
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A boy on a skateboard is pushed down the sidewalk by a friend. The skateboard rolls for a while but eventually comes to a stop.

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Friction is causing the skateboard to stop rolling.

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