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jok3333 [9.3K]
3 years ago
12

A race car is travelling around a circular track. The velocity of the car is 20 m/s, the radius of the track is 300 m, and the m

as of the car is 2000 kg. Determine the minimum coefficient of friction between the road surface and the tires of the car in order that the car does not skid off the track.
Engineering
1 answer:
Zielflug [23.3K]3 years ago
5 0

Answer:

μ = 0.136

Explanation:

given,

velocity of the car = 20 m/s

radius of the track = 300 m

mass of the car = 2000 kg

centrifugal force

F_c = \dfrac{mv^2}{r}

F_c = \dfrac{2000\times 20^2}{300}

F c = 2666. 67 N

F f= μ N

F f = μ m g

2666.67  =  μ × 2000 × 9.8

μ = 0.136

so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136

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Answer:

yes.

Explanation:

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3 years ago
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 16.403
Dmitry [639]

Answer:

A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

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A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol
insens350 [35]

<u>Explanation:</u>

Determine the weight of footing

W_{f}=\gamma(L)(B)(D)

Where W_{f} is the weight of footing, γ is the unit weight of concrete,  L is the length of footing is the width of footing, and D is the depth of footing

Substitute 2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3} for γ in the equation

\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

\sigma_{z p}^{\prime}=\gamma(D+B)-u

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute 18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0 for u in the equation

\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

\sigma_{z D}^{\prime}=\gamma D

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute \(18 kN / m ^{3}\) for \(\gamma, 0.5 m\) for \(D\) and 0 for \(u\) in the equation.

\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}

Here I_{e p} is the influence factor at the midpoint of soil layer  \sigma_{z^{p}}^{\prime} is initial vertical stress, \sigma_{z^{p}}^{\prime} is vertical effective stress, and Q is bearing pressure

Substitute 36 kPa for \(\sigma_{z p}^{\prime}, 228.47\) kPa for \(q,\) and 9 kPa for \(\sigma_{z D}^{\prime}\) in the equation\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}

Therefore the influence factor at the midpoint of the soil layer is 0.693

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Answer:

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Read 2 more answers
Let’s define a new language called dog-ish. A word is in the lan- guage dog-ish if the word contains the letters ’d’, ’o’, ’g’ a
attashe74 [19]

Answer and Explanation:

// code

class Main {

   public static void main(String[] args) {

       /*

        *

        *

        * your code

        *

        */

       System.out.println(inDogish("aplderogad"));

       System.out.println(inXish("aplderogad", "dog"));

   }

   // returns true if the word is in dog-ish

   // returns false if word is not in dog-ish

   public static boolean inDogish(String word) {

       // first find d

       if (dogishHelper(word, 'd')) {

           // first find string after d

           String temp = word.substring(word.indexOf("d"));

           // find o

           if (dogishHelper(temp, 'o')) {

               // find string after o

               temp = temp.substring(temp.indexOf("o"));

               // find g

               if (dogishHelper(temp, 'g'))

                   return true;

           }

The output is attached below

       }

       return false;

   }

   // necessary to implement inDogish recursively

   public static boolean dogishHelper(String word, char letter) {

       // end of string

       if (word.length() == 0)

           return false;

       // letter found

       if (word.charAt(0) == letter)

           return true;

       // search in next index

       return dogishHelper(word.substring(1), letter);

   }

   // a generalized version of the inDogish method

   public static boolean inXish(String word, String x) {

       if (x.length() == 0)

           return true;

       if (word.length() == 0)

           return false;

       if (word.charAt(0) == x.charAt(0))

           return inXish(word.substring(1), x.substring(1));

       return inXish(word.substring(1), x.substring(0));

   }

}

PS E:\fixer> java Main true true ne on

PS E:\fixer> java Main true true ne on

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3 years ago
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