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inn [45]
3 years ago
11

An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas

s of 1260 kg. The pavement is wet and oily, so the coefficient of kinetic friction between the car's tires and the pavement is only 0.500.
Physics
1 answer:
Viefleur [7K]3 years ago
4 0

1) Available force of friction: 6174 N

2) No

Explanation:

1)

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

F_f=\mu mg

where

\mu is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

\mu=0.500 (coefficient of friction)

m = 1260 kg (mass of the car)

g=9.8 m/s^2

Therefore, the force of friction is

F_f=(0.500)(1260)(9.8)=6174 N

2)

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

F=m\frac{v^2}{r}

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

v=54.1 km/h =15.0 m/s is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

F=(1260)\frac{15.0^2}{41.6}=6814 N

Therefore, the force of friction is not enough to keep the car in the curve, since F_f

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To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

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where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

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Replacing the previous equation with our values we have,

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9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

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The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

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Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

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That is equal in revolution to

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x = \theta*(2\pi*r)

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Explanation:

One period means time taken to complete one revolution.

In case of swings in one period time it travels the same position through two times.

Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.

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Answer:

When air resistance equals the weight of an object, the object has reached free fall.

Explanation:

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