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solmaris [256]
3 years ago
6

Explain the difference between the three types of friction

Physics
1 answer:
USPshnik [31]3 years ago
8 0
Static Friction

It is the friction that exists between a stationary object and the surface on which it's resting.


Sliding friction

It is the resistance created by two objects sliding against each other.

Rolling friction:-

It is the force resisting the motion when a body rolls on a surface.


hope this helps x
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A solid is 5 cm tall, 3 cm wide, and 2 cm thick. It has a mass of 129 g. What is its<br> density?
rjkz [21]

Answer:

4.3 g/cm³ or 4.3g/cc

Explanation:

Volume(V) = Height × Length × Width

= 5cm × 3cm × 2cm

= 30cm³

Mass(m) = 129gram

So,

Density = m/V

= 129g/30cm³

= 4.3g/cc or 4.3g/cm³

7 0
3 years ago
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
liq [111]

Answer:

Minimum time interval (t2)=0.90 SECONDS

Explanation:

  • coefficient of friction for employees footwear = 0.5
  • coefficient of friction for typical athletic shoe = 0.810
  • frictional force = coefficient of friction X acceleration due to gravity X mass of body
  • Acceleration due to gravity is a constant = 9.81 m/s
  • Let frictional force for employee footwear = FF1
  • Let frictional force for athletic footwear =FF2

                 FF1 = O.5 X 9.81 X mass of body

                         = 4.905 x mass of body

                  FF2 = 0.810 X 9.81 X mass of body

                          = 7.9461 x mass of body

The body started from rest there by making the initial velocity zero ( u = 0)

From d= ut + 1/2 a x t^{2}

  •      d = \frac{1}{2} x a x t^{2}  .....................................i  

            where d= distance and it is given as 3.25m

  •          F =ma  ...................................ii

making acceleration subject of the formula from equation ii

  •              a =\frac{F}{m}

         Making t subject of formula from equation (i)

  • t=\sqrt{\frac{2d}{(f/m} }

    where

  • \frac{FF1}{Mass of body} = 4.905
  • \frac{FF2}{Mass of body} =7.9461

  Let

  •            t1 = minimum time taken for frictional force for employee foot wear
  •                                 t1 = \sqrt{\frac{6.5}{4.905} } =1.15 seconds

  •                                  t2 = \sqrt{\frac{6.5}{7.9461} } = 0.90 seconds

 

THANK YOU

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