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IgorLugansk [536]
3 years ago
6

Our textbook describes a common model for friction (both static and kinetic) in which the area of contact between interacting su

rfaces is irrelevant to the size of the frictional force. However, we can't help noticing how wide the tires are on drag racing cars! One explanation for the large size of the tires would be to give a larger contact patch between the tire and the track in search of more traction. But our model says contact area doesn't matter. Clarify this situation in a mini essay of 2 or 3 sentences. Sort out whether contact area matters or not. Can you think of other advantages of really wide tires -- in addition to how cool they look.
Physics
1 answer:
Mkey [24]3 years ago
7 0

Contact area is not a factor of the amount of friction force acting between two bodies in contact, preventing rolling or sliding

Reasons:

  • Friction force is the product of the normal reaction and the coefficient of friction.
  • The coefficient of friction is higher in softer tires than hard tires which are required to be wider to reduce the pressure that can cause deformation withstood by hard tires.
  • The wide tires in drag racing cars can be thought of being made from the soft compound tire material having a higher coefficient of friction that increases the friction force.

Other advantages of wide tires are;

  • Increased traction
  • Shorter stopping distance

Learn more here:

brainly.com/question/17209968

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A diamond with a mass of 45 g hangs motionless from a chain. what is the upward force of the chain on the diamond?
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The upward force of the chain on the diamond would be the tension in the chain, and this tension would have to support the weight of the 45g that hangs from the chain.

mass = 45 g = 45/1000 kg = 0.045kg

Weight = mg = 0.045 * 10 ≈ 0.45N,            g ≈ 10 m/s²

So the upward force is ≈ 0.45N. 
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What acceleration would a 18 kg box have if we pushed it with a force of 250 N across a surface that
konstantin123 [22]

Answer:

a=0.08\ m/s^2

Explanation:

Given that,

Mass of a box, m = 18 kg

It is pushed with a force of 250 N

The coefficient of friction is 0.8

We need to find the acceleration of the box.

Normal force acting on the box,

N = mg

M = 18 × 10

N = 180 N

The force acting on the force in terms of coefficient of friction is given by :

F=\mu mg\\\\\text{or}\\\\ ma=\mu mg\\\\a=\dfrac{\mu }{g}\\\\a=\dfrac{0.8}{10}\\\\a=0.08\ m/s^2

So, the acceleration of the box is 0.08\ m/s^2

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Which interaction is most likely to result in the most competition among the organisms?
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A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and a
masya89 [10]

Answer:

1

  a =  2.82 \  m/s^2

2

 t =  8.87 \  s

3

 s =  204 \  m

Explanation:

From the question we are told that

   The  speed of the car is  v  =  25.0 \  m/s

   The length of the ramp is  d =  111 \ m

   The  constant velocity of the traffic v_t  =  23.0 \  m/s

Generally the acceleration of the car is mathematically represented as

           a =  \frac{v^2  -  u^2 }{2d}

Here  u is  equal to zero given that the car started from rest so

           a =  \frac{25^2 - 0^2 }{2 *  111}

       =>    a =  2.82 \  m/s^2

Generally the time taken is mathematically represented as

      t = \frac{ v - u}{ a}

=>   t  = \frac{ 25 - 0}{2.82}

=>   t =  8.87 \  s

The  distance traveled by the traffic is mathematically represented as

  s =  v_{t}t + \frac{1}{2} a t^2

Here  a is zero given that the traffic was moving at constant speed

=>    s  =  23 *  8.87

=>     s =  204 \  m

7 0
4 years ago
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