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Anna007 [38]
3 years ago
8

Can anyone explain how to do this to me? It is due tomorrow at 9:30am. Thanks.

Physics
1 answer:
Georgia [21]3 years ago
3 0
The 102N acting on the ropes being pulled by eric and kim have some of that force acting horizontally, and some of it vertically. By visualizing it as a right angled triangle, with the hypotenuse the length of the diagonal force, and each side the length of the horizontal and vertical forces, you can use trigonometry to calculate the length of the vertical force. You are told that it is at an angle of 30 with the vertical rope, therefore you know the length of the hypotenuse, and the angle between it and the vertical force, so using trig: (vertical force=x)
x/102=cos(30)
x=102*cos(30)
x=88.33
Therefore the diagonal ropes give a vertical force of 88.33N, and the centre rope, as it acts vertically, gives a vertical force of all 102N. The total:
88.33*2+102=278.66N

I don't know if this is very clear, I hope its good enough to help. If you don't understand, just ask, and I can answer any questions!!! :)
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If the frequency of a FM wave is 8.85 × 107 hertz, what is the period of the FM wave?
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Answer:

1.1299 x 10^-8   second    

Explanation:

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A movement that keeps the distal end of the body segment fixed in one location describes what type of kinetic chain movement? Op
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A movement which keeps the distal end of the body segment fixed in one location describes what type of kinetic chain movement is known as a <u>closed kinetic chain movement.</u>

<h3>What is a closed kinetic chain movement?</h3>

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The owner of a window treatment company wants to design Shades that will cover the windows of the house. the shades should also
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Para que ángulo el alcance es igual a la altura máxima en un lanzamiento de proyectil ideal sobre un plano horizontal. Translati
Gemiola [76]

Answer:

<h2>76°</h2>

Explanation:

In projectile,

Maximum height = \frac{u^{2}sin^{2} \theta  }{2g} and range = \frac{u^{2}sin2\theta }{g}

u = velocity of the body

\theta = angle of launch

g = acceleration due to gravity

If the range equals the maximum height in an ideal projectile launch on a horizontal plane, this means;

\frac{u^{2}sin^{2} \theta  }{2g} = \frac{u^{2}sin2\theta  }{g}

since sin2\theta = 2sin\theta cos\theta

\frac{u^{2}sin^{2} \theta  }{2g} = \frac{u^{2}2sin\theta cos\theta }{g}\\\frac{sin\theta}{2} =2cos\theta \\sin\theta = 4cos\theta\\\frac{sin\theta}{cos\theta}  = 4\\tan\theta = 4\\\theta = tan^{-1} \\4\theta = 75.9^{0}

The angle at which the range equals the maximum height in an ideal projectile launch on a horizontal plane is approximately 76°

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