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Akimi4 [234]
3 years ago
13

The driver of a car traveling up a 2% grade at an initial speed V0 applied the brakes abruptly and the vehicle slid to a complet

e stop at an average deceleration of 8 ft/s^2.
Was the pavement wet or dry?
Engineering
1 answer:
mafiozo [28]3 years ago
3 0

Answer:

The pavement was dry.

Explanation:

The car is modelled by means of the equations of equilibrium: (x' is for the axis parallel to the incline, y' is for the axis perpendicular to the incline)

\Sigma F_{x'} = -\mu_{k}\cdot N-m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N-m\cdot g \cos \theta = 0

After some algebraic handling, the following expression is constructed:

-\mu_{k}\cdot m \cdot g \cdot \cos \theta - m \cdot g \cdot \sin \theta = m \cdot a

-g\cdot (\mu_{k} \cdot \cos \theta +\sin \theta) = a

\mu_{k} \cdot \cos \theta = -\frac{a}{g} - \sin \theta

\mu_{k} = -\frac{1}{\cos \theta}\cdot \left(\frac{a}{g} +\sin \theta \right)

The angle of the incline is:

\theta = \tan^{-1} 0.02

\theta \approx 1.146^{\textdegree}

Now, the kinetic coefficient of friction is:

\mu_{k} = -\frac{1}{\cos 1.146^{\textdegree}}\cdot \left(-\frac{8\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }+\sin 1.146^{\textdegree}\right)

\mu_{k}\approx 0.796

A typical kinetic coeffcient of friction between a car tire and asphalt is about 0.6, if pavement would be wet, such indicator would be significantly lower. Therefore, the deceleration occurs on dry pavement.

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) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p
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Answer:

number of pulses produced =  162 pulses

Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

linear displacement = 200 mm

solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

circumference = 2 × π × 50

circumference = 314.16 mm

so now we get number of pulses produced

number of pulses produced = \frac{linear\ displacement}{circumference} × No of pulses per revolution .................2

number of pulses produced = \frac{200}{314.16} × 256

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5 0
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Which of the following has led to a safer and more prosperous global community within the last century? the Bronze Age composite
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There are two types of cellular phones, handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles. P
nexus9112 [7]

Answer:

A) P(W) = 0.5

B) P(MF) = 0.3

C) P(H) = 0.6

Explanation:

We are told that there are two types of cellular phones which are handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles.

Also, Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W).

Thus, the sample space is combination of types and classification we are given and it is written as;

S = {HF, HW, MF, MW}

A) Now, phones can either be fast(F) or slow(W). Thus, we can write;

P(F) + P(W) = 1

We are given P(F) = 0.5

Thus;

0.5 + P(W) = 1

P(W) = 1 - 0.5

P(W) = 0.5

B) Now, from the problem statement, a phone call can either be made with a handheld(H) or mobile(M). Thus the sample space partition is {H, M} and we can express as;

P(H ∩ F) + P(M ∩ F) = P(F)

We are given P[F] = 0.5 and P[HF] = 0.2.

P(H ∩ F) is same as P[HF]

Also, P(M ∩ F) is same as P(MF)

Thus;

0.2 + P(MF) = 0.5

P(MF) = 0.5 - 0.2

P(MF) = 0.3

C) Similarly, mobile Phone calls can either be fast or slow. It means the sample space partition is {F, W}

Thus;

P(M) = P(MW) + P(MF)

P(M) = 0.1 + 0.3

P(M) = 0.4

Now, since cellular phones can either be handheld(H) or Mobile(M), then we can say;

P(H) + P(M) = 1

P(H) + 0.4 = 1

P(H) = 1 - 0.4

P(H) = 0.6

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