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adell
2 months ago
10

To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th

e pole after 0.50 s and then reaches the top of the pole again after a total elapsed time of 4.1 s. How high is the pole above the point where the ball was launched? (You can ignore air resistance.) To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of the pole after 0.50 s and then reaches the top of the pole again after a total elapsed time of 4.1 s. How high is the pole above the point where the ball was launched? (You can ignore air resistance.) 16 m 13 m 18 m 26 m 10 m
Physics
1 answer:
Softa [3K]2 months ago
5 0

Answer:

H = 10.05 m

Explanation:

The stone reaches the top of the flagpole at both t = 0.5 s and t = 4.1 s

therefore, the total duration of the upwards motion above the peak of the pole is provided as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

this indicates the speed at the flagpole's top

at this point we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

the height of the flagpole is stated as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

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Answer:

Statements 4, 6 & 7 are incorrect.

Explanation:

In any elastic collision, the overall momentum vector sum of the system remains zero.

In this scenario, an elastic collision occurs between the ball and a stationary wall. The ball's velocity will consistently revert after the impact, leading to a change in direction of momentum.

The initial momentum of the ball is represented as:

p=m.v

where:

m = mass of the ball

v = initial velocity of the body

post-collision for the elastic interaction:

p=m.(-v)

  • Here, the momentum changes solely in direction, thus contradicting statement 7.
  • During the impact, both the ball and the wall exert forces on each other that are equal and opposite. The wall remains motionless, while the ball is influenced by the wall's reaction force, performing work on it, which contradicts statement 4.
  • Given that this collision is elastic, the ball's form and dimensions do not alter.
  • The previous points clearly indicate that not all provided statements hold true, thus violating statement 6.
4 0
28 days ago
2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind
Yuliya22 [3333]

The speed resulting from the plane is (3) 226 m/s

Reasoning:

We can determine the plane's resultant speed using the Pythagorean Theorem because the two speeds form a right angle (creating a right triangle).

Thus, the computation is as follows:

ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}

ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}

Consequently, the plane's resultant speed is (3) 226 m/s

Have a wonderful day!

5 0
1 month ago
two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb dire
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Thanks for asking your question here. I hope this response provides clarity. Feel free to ask additional questions. The moment resulting from the two forces about point O is 376 lb-ft counterclockwise.
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A champion athlete can produce one horsepower (746 W) for a short period of time. The number of 16-cm-high steps a 70-kg athlete
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Answer:

407 steps

Explanation:

Based on the question,

P = mgh/t........... Equation 1

Where P stands for power, m is mass, g denotes gravity, h is height, and t represents time.

Rearranging the equation to solve for h, we have:

h = Pt/mg............. Equation 2

Providing values: P = 746 W, t = 1 minute = 60 seconds, m = 70 kg.

Given constant: g = 9.8 m/s²

By substituting into equation 2

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Calculating number of steps: 6525/16

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Use this free body diagram to help you find the magnitude of the force F2 needed to keep this block in static equilibrium. WILL
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