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nirvana33 [79]
3 years ago
10

On an essentially frictionless, horizontal ice rink, a skater moving at 5.0 m/s encounters a rough patch that reduces her speed

by 40 % due to a friction force that is 22 % of her weight. Use the work-energy theorem to find the length of this rough patch. Express your answer using two significant figures.
Physics
1 answer:
Leviafan [203]3 years ago
3 0

Answer:  the length of the rough patch is 4.9 m.

Explanation:

The work - energy theorem, in simple words, expresses that the change in the kinetic energy in an object, is equal to the work done on the object by non-conservative forces, like frictional ones.

In the question, we know that the skater is moving at a given speed, and due to the effects of friction, her speed is reduced to 40%.

Added to this, we are told that the friction force is equal to 22% of her weight so we can write the following:

Ff = 0.22 m. g.

Now the work done by this force, is equal to the product of the force times the distance during which the force acted, i.e,. the distance that we are looking for.

So, we can write the following in a brief:

ΔK = Wf ⇒ 1/2 m (vf² - v₀²) = -Ff . x

(The negative sign explains that the friction force always opposes to the relative movement between the two surfaces in contact).

Replacing by the values, and solving for x, we get:

1/2 ( 2² - 5²) = -0.22. 9.8. x   ⇒  x = 4.9 m

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3 years ago
What is the highest degrees above the horizon the moon ever gets during the year in the Yakima Valley ?
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The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.

Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima.  The latitude in the middle
of that intersection is 46.585° North.  <u>That's</u> the number we need.

Here's how I would do it:

-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.

-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°. 

-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.

This sets the limit of the highest in the sky that the moon can ever appear.

90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .

That doesn't happen regularly.  It would depend on everything coming
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Depending on the time of year, that can be any time of the day or night.

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67° above the horizon at midnight.


5 0
3 years ago
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