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lesantik
9 days ago
12

A gazebo is being built near a nature trail. An equation of the line representing the nature trail is y = 13x−4. Each unit in th

e coordinate plane corresponds to 10 feet. Approximately how far is the gazebo from the nature trail? Round your answer to the nearest foot.
Mathematics
1 answer:
PIT_PIT [9.1K]9 days ago
6 0

Respuesta:

126 pies

Explicación paso a paso:

Dado que:

Ecuación de la línea que representa el sendero de naturaleza:

y = 13x - 4

Distancia del gazebo respecto al sendero de naturaleza:

Cada unidad en el plano de coordenadas = 10 pies

Por lo tanto,

y = 13(10) - 4

y = 130 - 4

y = 126 pies

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Detailed explanation

This scenario involves motion with constant acceleration.

The relevant variables include the following.

\boxed{u \ or \ v_i = initial \ velocity}

\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}

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We know the flea attains a takeoff velocity of 1.0 m/s over a distance of 0.50 mm.

The flea starts from rest, so the initial velocity is zero. The question asks for the flea's acceleration during leg extension.

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\boxed{ \ v^2 = u^2 + 2ad \ }

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First, convert 0.50 mm to meters: \boxed{0.50 \ mm = 0.50 \times 10^{-3} \ m = 5.0 \times 10^{-4} \ m}

Solution steps:

Rearrange the formula to isolate acceleration (a).

\boxed{ \ v^2 = u^2 + 2ad \ }

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Insert the given values into the rearranged equation.

\boxed{ \ a = \frac{(1.0)^2 - (0)^2}{2(5.0 \times 10^{-4})} \ }

\boxed{ \ a = \frac{1}{10 \times 10^{-4}} \ }

\boxed{ \ a = \frac{1}{1 \times 10^{-3}} \ }

\boxed{ \ a = 1 \times 10^{3}\ = 1,000 \ m/s^2 \ }

This yields the flea's acceleration as 1,000 m/s².

Additional resources

  1. Scientific notation: brainly.com/question/7263463
  2. Determining substance mass: brainly.com/question/4053884
  3. Conversion into cubic units: brainly.com/question/1446243

Keywords: flea, jumping, leg extension, takeoff velocity, initial speed, displacement, acceleration, constant acceleration, unit conversion

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