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sveticcg [70]
3 years ago
10

A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 34 m/s when a 60

kg skydiver drops out by releasing his grip on the glider. What is the gliders speed just after the skydiver lets go? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
german3 years ago
3 0

Answer:

34 m/s

Explanation:

m = Mass of glider with person = 680 kg

v = Velocity of glider with person = 34 m/s

m_1 = Mass of glider without person = 680-60 kg

v_1 = Gliders speed just after the skydiver lets go

m_2 = Mass of person = 60 kg

v_2 = Velcotiy of person = 34 m/s

As the linear momentum of the system is conserved

m_1v_1+m_2v_2=mv\\\Rightarrow v_1=\dfrac{mv-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{680\times 34-60\times 34}{680-60}\\\Rightarrow v_1=34\ m/s

The gliders speed just after the skydiver lets go is 34 m/s

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You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of
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Answer:

P = 1.090\,N

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The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

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Now, the exerted force is:

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Lastly, the force is calculated:

P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )

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A train travels 8.81 m/s in a -51.0° direction.
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The displacement of the train after 2.23 seconds is 25.4 m.

<h3>Resultant velocity of the train</h3>

The resultant velocity of the train is calculated as follows;

R² = vi² + vf² - 2vivf cos(θ)

where;

  • θ is the angle between the velocity = (90 - 51) + 37 = 76⁰

R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)

R² = 129.75

R = √129.75

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<h3>Displacement of the train</h3>

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Δx = 11.39 m/s x 2.23 s

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Thus, the displacement of the train after 2.23 seconds is 25.4 m.

Learn more about displacement here: brainly.com/question/2109763

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