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Lostsunrise [7]
3 years ago
8

What are some things cyclists do to reduce friction?

Physics
1 answer:
Juli2301 [7.4K]3 years ago
6 0

1. Cyclists could ride behind other cyclists to reduce the overall effects of wind resistance

2. Friction = μN --> where N is the normal force and μ is the coefficient of static or dynamic friction. By reducing the weight of the bike the could reduce the reaction force, which is the normal force that acts on them, which is friction is dependent on.

3. Also by getting a smother tire, which reduces the friction between the ground and the tire. This is why professional cyclists have smooth and thin tires meant for speed while mountain bikes generally have large projects called the tread of the tire, to increase friction so they dont slip while biking on rough terrain.

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A human hair is approximately 50 pm in diameter. Express this diameter in meters.
cupoosta [38]

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4 0
3 years ago
a man hikes 6.6 km north along a straight path with an average velocity of 4.2 km/h to the north. he rest at a bench for 15 min.
SSSSS [86.1K]

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2.6h

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I attached the image below of the work hope you can see it. Hope this helps!

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A person jumps from a plane in is falling the person releases a parish you in continues to fall the person lays 35.2 kg and ther
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7 0
2 years ago
Please help me
qaws [65]

-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.

-- We know that the y-component of velocity is the derivative of the
y-component of position.

-- We're given the y-component of position as a function of time.

So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.

Now, the position function may look big and ugly in the picture.  But with the
exception of  't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation.  The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.

From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶

First derivative . . . y' (t) = (a₀ - g) t  -  6 (a₀ / 30t₀⁴ ) t⁵  =  (a₀ - g) t  -  (a₀ / 5t₀⁴ ) t⁵

There's your velocity . . . /\ .

Second derivative . . . y'' (t) = (a₀ - g) -  5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) -  (a₀ /t₀⁴ ) t⁴

and there's your acceleration . . . /\ .
That's the one you're supposed to graph.

a₀ is the acceleration due to the model rocket engine thrust
     combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀  is how long the model rocket engine burns

Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.

The big name in model rocketry is Estes.  Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.


6 0
3 years ago
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