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Lostsunrise [7]
4 years ago
8

What are some things cyclists do to reduce friction?

Physics
1 answer:
Juli2301 [7.4K]4 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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Pls answer quick I need to get the answer rn
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I think it is False because as the Gad relajases fuel it doesn’t move as much anymore
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3 years ago
Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip’s speed at impact
Westkost [7]

Answer:

-5.8868501529 m/s² or -5.8868501529g

0.118909090909 s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{1.3^2-2^2}{2\times 0.02}\\\Rightarrow a=-5.8868501529\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{-57.75}{9.81}\\\Rightarrow a=-5.8868501529g

The acceleration is -5.8868501529 m/s² or -5.8868501529g

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{1.3-2}{-5.8868501529}\\\Rightarrow t=0.118909090909\ s

The time taken is 0.118909090909 s

7 0
3 years ago
A space with an absolute pressure less than one atmosphere may be considered?
jenyasd209 [6]
It considered as Zero Gage pressure. 
3 0
3 years ago
An object moving at 38m/s takes 4s to come to a stop. What is the object’s acceleration?
Rudik [331]
- 9.5 m/s^2

use the SUVAT method 

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6 0
4 years ago
A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is th
vladimir2022 [97]

As we know that friction force on box is given by

F_s = \mu_s N

here we know that

N = mg

here we have

m = 12 kg

\mu_s = 0.42

so now we have

N = 12(9.8) = 117.6 N

now we will have

F_s = 0.42(12)(9.8)

F_s = 49.4 N

so it required minimum 49 N(approx) force to move the block

5 0
4 years ago
Read 2 more answers
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