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3 years ago

An 80kg student running at 3.50m/s grabs a rope that is hanging vertically. How high will the student swing?

Physics

1 answer:

Rudiy273 years ago

8 0

Answer:

The student will swing to a height of 61.25 cm.

Explanation:

Given that,

Mass of the student, m = 80 kg

Speed of student, v = 3.5 m/s

The student running at this speed grabs a rope that is hanging vertically. In the whole process, the energy of the system remains conserved. So, using conservation of energy to find the height reached by the student. So,

$\dfrac{1}{2}mv^2=mgh$

h is height reached by the student

$h=\dfrac{v^2}{2g}\\\\h=\dfrac{(3.5)^2}{2\times 10}\\\\h=0.6125\ m\\\\h=61.25\ cm$

So, the student will swing to a height of 61.25 cm.

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6. The image to the right shows a moment of inertia

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2 years ago

During a very quick stop, a car decelerates at 7.6 m/s2. Assume the forward motion of the car corresponds to a positive directio

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Answer:

24.57 revolutions

Explanation:

(a) If they do not slip on the pavement, then the angular acceleration is

$\alpha = a / r = 7.6 / 0.26 = 29.23 rad/s^2$

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$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where v = 0 m/s is the final angular velocity of the wheel when it stops, $\omega_0$ = 95rad/s is the initial angular velocity of the wheel, $\alpha = -29.23 rad/s^2$ is the deceleration of the wheel, and $\Delta \theta$ is the angle swept in rad, which we care looking for:

$0 - 95^2 = 2*29.23\Delta \theta$

$9025 = 58.46 \Delta \theta$

$\Delta \theta = 9025 / 58.46 = 154.375 rad$

As each revolution equals to 2π, the total revolution it makes before stop is

154.375 / 2π = 24.57 revolutions

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