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ELEN [110]
3 years ago
10

Find the magnitude of a fourth force on the stone that will make the vector sum of the four forces zero.

Physics
1 answer:
Tanya [424]3 years ago
7 0
100cos30° + 80cos120° + 40cos233° + Dx = 0 
<span>Dx = -22.53 N </span>

<span>y components: </span>
<span>100sin30° + 80sin120° + 40sin233° + Dy = 0 </span>
<span>Dy = -87.34 N </span>

<span>magnitude of D: </span>
<span>sqrt[(-22.53)² + (-87.34)²] </span>
<span>90.2 N </span>

<span>direction of D: </span>
<span>arctan[(-87.34)/(-22.53)] </span>
<span>75.5° ref, but since Dx and Dy are both negative, we know this vector is in QIV: </span>
<span>360 - 75.5° = 284.5°</span>
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Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton
pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx  

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:


F = K * \Delta{x}

2\:N = 4\:N/cm*\Delta{x}

4\:N/cm*\Delta{x} = 2\:N

\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}

simplify by 2

\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}

\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark

Answer:

B.) 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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