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disa [49]
3 years ago
9

Numeria

Physics
2 answers:
Vinil7 [7]3 years ago
5 0

Answer:

hope it helps

plz mark as brainliest

polet [3.4K]3 years ago
3 0

Answer:

<h3>\boxed{40 \: cm}</h3>

Explanation:

Load ( L ) = 800 N

Effort ( E ) = 200 N

Load distance ( LD ) = 10 cm

Effort distance ( ED ) = ?

now, Let's find the effort distance:

\mathsf{L \times LD = E \times ED}

Plug the values

\mathsf{800  \times 10 = 200 \times ED}

Multiply the numbers

\mathsf{8000 = 200 \: ED}

Swipe the sides of the equation

\mathsf{200 \: ED = 8000}

Divide both sides of the equation by 200

\mathsf{ \frac{200 \: ED }{200}  =   \frac{8000}{200} }

Calculate

\mathsf{ED  \:  =  \: 40 \: cm}

Hope I helped!

Best regards!

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Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the
Gnesinka [82]

Answer:

F = 233.52 N,  θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

        = 80.5 N

Jill

       cos 45 = F_{2x} / F₂2

       sin 45 = F_{2y} / F₂2

       F_{2x} = F₂ cos 45

       F_{2y} = F₂ sin 45

       F_{2x} = 81.7 cos 45 = 57.77 N

       F_{2y} = 81.7 sin 45 = 57.77 N

Jane

      cos (270 + 45) = F_{3x} / F₃3

      sin 315 = F_{3y} / F₃

      F_{3x} = 131 cos 315 = 92.63 N

      F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

         F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

        F = √(230.9² + 34.86²)

        F = 233.52 N

let's use trigonometry for the angle

        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

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