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Alenkasestr [34]
3 years ago
14

28. Someone throws a rubber ball vertically upward from the roof of a building 8.00 m in height. The ball rises,

Physics
1 answer:
Debora [2.8K]3 years ago
6 0

Answer:

v_o= 12 m/s

Explanation:

The ball experienced a constant acceleration motion, so we need to apply the following formula:

y=y_o+v_o*t+\frac{1}{2}*a*t^2\\\\vo=\frac{y-y_o-\frac{1}{2}*a*t^2}{t}\\\\v_o=\frac{0m-8m-\frac{1}{2}*(-9.81m/s^2)*(3.00s)^2}{3.00s}\\v_o=12m/s

Note: we set the acceleration of gravity as negative because it is going down.

the ball was thrown vertically upward with an initial velocity of 12 m/s

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The value of the force, F₀, at equilibrium is equal to the horizontal

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Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

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M·g = T₂·cos(37°) + T₁

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Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

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