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igor_vitrenko [27]
3 years ago
11

Can someone help me asap please​

Physics
1 answer:
Vesna [10]3 years ago
4 0

Answer:

<u>Resultant</u><u> </u><u>force</u><u> </u><u>is</u><u> </u><u>2</u><u>0</u><u> </u><u>N</u>

Explanation:

Resolving vertically:

{ \sf{F_{y} = (8 \cos 30 \degree) + (12 \cos 30 \degree)   }} \\ { \sf{ \sum F_{y} = 17.3 \: newtons }}

Resolving horizontally:

{ \sf{F_{x} = (8 \sin 30 \degree) + (12 \sin 30 \degree)   }} \\ { \sf{ \sum F _{x}  = 10 \: newtons}}

Resultant force:

{ \boxed{ \bf{F =  \sqrt{ {F _{x} }^{2} +   {F _{y} }^{2} } }}} \\ { \sf{F =  \sqrt{ {17.3}^{2} +  {10}^{2}  } }} \\ { \sf{F =  \sqrt{399.29} }} \\ F = 19.98 \approx20  \: newtons

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Answer:

  v₀ = 6.64 m / s

Explanation:

This is a projectile throwing exercise

          x = v₀ₓ t

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In this case they indicate that y₀ = 1.8 m and the point of the basket is x=4.9m y = 3.0 m

         

the time to reach the basket is

        t = x / v₀ₓ

we substitute

        y- y₀ = \frac{ v_o \ x \ sin \theta  }{ v_o \ cos \theta} - \frac{1}{2} g \ \frac{x^2 }{v_o^2 \ cos^2 \theta }

        y - y₀ = x tan θ - \frac{ g \ x^2 }{ 2 \ cos^2 \theta } \ \frac{1}{v_o^2 }

         

we substitute the values

        3 -1.8 = 3.0 tan 60 - \frac{ 9.8 \ 3^2 }{2 \ cos^2 60 } \ \frac{1}{v_o^2}

        1.2 = 5.196 - 176.4 1 / v₀²

        176.4 1 / v₀² = 3.996

        v₀ = \sqrt{ \frac{ 176.4}{3.996} }

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3 years ago
Vector A and B are given as follows:
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Answers

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Answer:

149,916J

Explanation:

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