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kenny6666 [7]
2 years ago
10

Ben walks 500 meters from his house to the corner store. He then walks back toward his house but stops to talk to a neighbor whe

n he is 200 meters from the corner store. If it takes Ben 910 seconds from the time he leaves his house until he stops to talk to his neighbor, what is his average velocity?
Physics
2 answers:
spin [16.1K]2 years ago
7 0
Average velocity =

      (displacement) / (time for the displacement)
and
      (direction of the displacement) .

Displacement =

      (distance from the start-point to the end-point)
and
      (direction from the start-point to the end-point)   .

When Ben is 200 meters from the corner store,
he is (500 - 200) = 300 meters from his house.

His displacement is

         300 meters in the direction
                             from his house to the neighbor .


His average velocity is

         (300/910) =  0.33 meters per second, in the
                               direction from his house to the neighbor .


Kay [80]2 years ago
7 0

Answer:

0.3 m/s

Explanation:

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The speed of the block when the compression is 15 cm is 9.85 m/s.

The given parameters;

  • <em>mass of the block, m = 2.4 kg</em>
  • <em>height of the block, h =  5 m</em>
  • <em>compression of the spring, x = 25 cm = 0.25 m</em>

The spring constant is calculated as follows;

F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2  - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2  = mgh -   \frac{1}{2} kx^2\\\\mv^2   = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.

Learn more here:brainly.com/question/14289286

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