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anygoal [31]
3 years ago
10

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction

between the road and the car's tires that will allow the car to travel atthis speed without sliding?
a. 1.23
b. 0.662
c. 0.816
d. 0.952
e. 0.736
Physics
1 answer:
zhenek [66]3 years ago
7 0

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, F_c=\frac{mv^2}{R}

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

f=\mu N

As there is no vertical motion, therefore, N=mg. So,

f=\mu mg

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816

Therefore, option (c) is correct.

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Which fractures tend to occur first when a pane of glass is struck?
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True or false. The greater the distance that the plane moves from an object, the lower the force that will be applied when the a
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A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. Its initial position
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Answer:

Follows are the solution to this question:

Explanation:

In point a:

Place of particles

X(t)=\int V_{x}(t)dt

       =\int 2t^{2}dt\\\\=\frac{2}{3}t^{3}+C

\to t=0\\\\ \to X(0)=2.3 \ m

\to X(0)=0+C\\\\ \to C=2.3\  m

\to X(t)=( \frac{2}{3})t^3 + 2.3\\\\ \to t=2.2\\\\\to X=( \frac{2}{3})\times 2.2^3 +2.3 \\\\

        = \frac{2}{3}\times 10.648 +2.3\\\\= \frac{21.296}{3}+2.3\\\\  = 7.09+2.3\\\\ =9.39\\\\ =9.4\ m

In point b:

when t=2.2 \ s

the Particle velocity  (V)=2 \times 2.22 =9.68\  \frac{m}{s}

In point c:

Calculating the Particle acceleration:

\to a=\frac{dV}{dt} =4\ t\\\\\to t=2.2 \ s\\\\\to a=4\times 2.2  =8.8 \ \frac{m}{s^2}

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