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Lady bird [3.3K]
3 years ago
8

A car slams on the brakes to stop. The road pushes back up on the car with a normal force of 12,750 N and friction from the brak

es provides 9,560 N of force to stop the car. Calculate the coefficient of friction between the brakes and the wheels.
Physics
1 answer:
ivann1987 [24]3 years ago
6 0

Answer:

\displaystyle \mu_d=0.75

Explanation:

Coefficients of Friction

Objects in physical contact produce friction which usually manifests as thermal energy being dissipated in the surface where the objects are interacting. It's usually harder to start to move an object from rest, that keeps moving it at a constant speed on the same surface. That is why there are two different coefficients of friction: the static and the dynamic. As mentioned, the static coefficient \mu_s is greater than the dynamic coefficient \mu_d. The car is already moving and is attempting to stop. The coefficient of friction is defined as

\displaystyle \mu_d=\frac{F_r}{N}

Where Fr is the force of friction and N is the normal or the force the road pushes back up on the car. With the given data, we have

\displaystyle \mu_d=\frac{9,560\ N}{12,750\ N}

\displaystyle \boxed{\mu_d=0.75}

The coefficient of friction is dimensionless (doesn't have any units)

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In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 350MV . Th
AleksAgata [21]

Complete Question

In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 350 MV (35,000,000 V). The bottoms of the thunderclouds are typically 1500 m above the earth, and can have an area of 120 km^2. Modeling the earth/cloud system as a huge capacitor, calculate

a. the capacitance of the earth-cloud system

b. the charge stored in the "capacitor"

c. the energy stored in the capacitor

Answer:

a

 C =  7.08 *10^{-7} \  F

b

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c

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Explanation:

From the question we are told that

The potential difference is  V  =  35000000 V

The distance of the bottom of the thunderstorm from the earth is  d = 1500 m

The area is  A =  120 \  km^2 =  120 *10^{6} \  m^2

Generally the capacitance of the earth cloud system is mathematically represented as

         C =  \epsilon_o *  \frac{A}{d}

 Here \epsilon_o is the permitivity of free space with as value \epsilon_o =  8.85 *10^{-12} \  C/(V\cdot m)

So

     C =  8.85*10^{-12} *  \frac{120*10^{6}}{1500}

=>  C =  7.08 *10^{-7} \  F

Generally the charge stored in the capacitor (earth-cloud system) is mathematically represented as

       Q =  C  *  V

=>    Q =  7.08 *10^{-7}  *   35000000

=>    Q =  24.78 \  C

Generally the energy stored in the capacitor is mathematically represented as

       E = \frac{1}{2}  * Q *  V

=>    E = \frac{1}{2}  *   24.78 *  35000000

=>    E =433650000 \ J

   

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