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Olenka [21]
3 years ago
9

what does it mean if the coefficient of friction is greater? for example,a coefficient of friction of 0.1 compared to 0.7

Physics
2 answers:
Paul [167]3 years ago
4 0
If the coefficient of a fraction is greater, than the overall fraction is greater than 1. In your example you misspelled fraction wrong, and gave an incorrect example which momentarily confused me, but 0.1/0.7 is a fraction where the numerator/coefficient is smaller than the one on the bottom, or the denominator. So the overall fraction is less than 1.
Hope this helps!
Sergeu [11.5K]3 years ago
3 0
Well, 0.1 is actually less than 0.7, but I understand what you're asking.

The coefficient of friction describes the relationship between two surfaces
that are sliding by each other.  The higher the coefficient of friction is, the
'rougher' the meeting is, and the harder it is for one to slide over the other. 
A skate blade against ice has a very low coefficient of friction.  Sandpaper
against blue jeans has a high coefficient of friction.

A higher coefficient of friction means that when one thing is sliding over
the other one, friction robs more energy from the motion.  It's harder to
push one thing over the other one, and when you let go, the moving one
slows down and stops sooner.

Air resistance is actually an example of friction.  It prevents falling things
from falling as fast as they would if there were no air. The coefficient of
friction when something moves through air is pretty low.  If the same
object were trying to move through molasses or honey, the coefficient
of friction would be greater.

Friction robs energy, and turns it into heat.  So, especially in machinery with
moving parts, we want to make the coefficient of friction between the moving parts
as small as possible.  That's what the OIL in a car's engine is for.
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A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera
Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m=2.1\times 10^3 kg

Initial velocity,u=0

Distance,s=5.9 m

\theta=19^{\circ}

Average friction force,f=4.0\times 10^3 N

We have to find the speed of the car at the bottom of the driveway.

Net force,F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3

Where g=9.8 m/s^2

Acceleration,a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}

v=\sqrt{2as}

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3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

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