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3 years ago

13

Look at the picture below. The boy is swinging back and forth on the swing. At which point is the potential energy of a swing th

e greatest?
Point A (top)

Point B (halfway down)

Point C (bottom)

Point D (halfway up)

2 answers:

Komok [63]3 years ago

5 0

The correct answer is Point A the top

kvv77 [185]3 years ago

3 0

Point A (top)

Potential energy is the energy possessed by an object due to its height relative to a reference level. so greater the height above the reference level, greater will be the potential energy.

when the boy is at the top, the height is maximum there. hence the potential energy of the boy is maximum at the Top.

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$v=\sqrt{2gh}$

So the speed at the lowest point is $v=7 m/s$

Now, using the conservation of momentum we have:

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Thank you for posting your question here at brainly. Feel free to ask more questions.
<span>The best and most correct answer among the choices provided by the question is B. Reaches a max height of 8.25 feet after 0.63 seconds</span> . <span><span>

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3 years ago

Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh

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Answer:

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Explanation:

Physically speaking, the resulting velocity of the helicopter ( $\vec v_{H}$ ), measured in meters per second, is equal to the absolute velocity of the wind ( $\vec v_{W}$ ), measured in meters per second, plus the velocity of the helicopter relative to wind ( $\vec v_{H/W}$ ), also call velocity at still air, measured in meters per second. That is:

$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

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$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

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$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

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$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

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