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igor_vitrenko [27]
3 years ago
13

A ball is thrown upward. As it passes 5.0 m height it is traveling at 4.0 m/s up. What was its initial upward velocity? (a) 7.0

m/s ( b) 9.1 m/s (c) 10.7 m/s (d) 12.1 m/s (e) 14.6 m/s
Physics
1 answer:
sveticcg [70]3 years ago
5 0

Answer:

c) 10.7m/s

Explanation:

From the exercise we know that at 5m the ball  is traveling at 4m/s

To calculate its initial velocity we need to solve the following equation:

v_{y}^{2}=v_{oy}^{2}+2g(y-y_{o})

Since the initial height is 0

Solving for v_{o}

v_{oy}=\sqrt{v_{y}^{2}-2gy}=\sqrt{(4m/s)^2-2(-9.8m/s^2)(5m)}=10.7m/s

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3 years ago
A person yells across a canyon and hears an echo 2.5 seconds later. The air temperature is 21°C. How far away is the canyon wall
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Answer: The canyon wall is 850 m away from the person

Explanation:

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V=\sqrt{\frac{(1.4)(8.314 J/mol.K)(294.15 K)}{0.029 kg/mol}}

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V=\frac{d}{t} (3)

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d is the distance between the person and the canyon wall

t=2.5 s is the time it takes to the sound wave to travel from the person and then go back

Isolating d:

d=V.t (4)

d=(343.60 m/s)(2.5 s) (5)

Finally:

d=850 m

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