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IgorLugansk [536]
2 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]2 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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The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

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Solution :

Given :

The distance between the two loud speakers, d = 1.8 \ m

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point D, the speakers are out of phase and so the path difference is $=\frac{\lambda}{2}$

Therefore,

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$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$

$\lambda = 2 \times 0.4985$

$\lambda = 0.99714 \ m$

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$f=\frac{v}{\lambda}$

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Answer:

The force is  F = 3920 \ N

Explanation:

The diagram for this question is shown on the first uploaded image

   From the question we are told that

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