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Alja [10]
3 years ago
6

Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalp

ost and experiences a backward force of 1.76×104 n for 5.50×10–2 s .
Physics
1 answer:
Contact [7]3 years ago
6 0
<span>The answer is -0.8 m/s. We know acceleration is the average of final minus initial velocity over time (a = (vf-v0)/t). We also know that Force is equal to Mass times Acceleration (F = ma). Using our force equation, we know that the acceleration we get is negative 8.8 (-8.8). The force is acting in the opposite direction of the rugby player, hence the negative sign. From there, plug in that number for a in the velocity equation, and solve for vf, as v0 and t are known. We get 0.8 m/s in the opposite direction that the player was running.</span>
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A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of
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Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height \Delta h depends on the square of the time:

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t,

where

  • \Delta h is the change in the ball's height.
  • g is the acceleration due to gravity.
  • t is the time for which the ball is in the air.
  • v_0 is the initial vertical velocity of the ball.
  • The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. \Delta h = -0.950\;\text{m}.
  • Gravity pulls objects toward the earth, so g is also negative. g \approx -9.81\;\text{m}\cdot\text{s}^{-2} near the surface of the earth.
  • Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, v_0 = 0.

Solve for t.

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t;

\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2};

\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)};

t \approx 0.440315\;\text{s}.

What's the initial horizontal velocity of the ball?

  • Horizontal displacement of the ball: \Delta x = 0.352\;\text{m};
  • Time taken: \Delta t = 0.440315\;\text{s}

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}.

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

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