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3 years ago

To test the performance of its tires, a car

Physics

1 answer:

Rom4ik [11]3 years ago

5 0

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

$\mu mg=m\frac{v^2}{r}$

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

$\mu$ is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

$g=9.8 m/s^2$

And re-arranging the equation for $\mu$ , we can find the coefficient of static friction:

$\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222$

Learn more about friction:

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Or speed is distance over time

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Non metals tend to have higher ionization energies

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A stuntman of mass 48 kg is to be launched horizontally out of a spring-

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The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.

<h3 /><h3>Velocity: </h3>

This is the ratio of displacement to time. The S.I unit of Velocity is m/s. The velocity of the stuntman can be calculated using the formula below.

⇒ Formula:

mv²/2 = ke²/2
mv² = ke².................. Equation 1

⇒ Where:

m = mass of the stuntman
v = velocity of the stuntman
k = force constant of the spring
e = compression of the spring

⇒ Make v the subject of the equation

v = √(ke²/m)................. Equation 2

From the question,

⇒ Given:

m = 48 kg
k = 75 N/m
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⇒ Substitute these values into equation 2

v = √[(75×4²)/48]
v = √25
v = 5 m/s.

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The right option is O A. 5 m/s

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3 years ago

The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

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Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$

Where:

$K_{1}$ - Kinetic energy at point 1, measured in joules.

$U_{1}$ - Gravitational potential energy at point 2, measured in joules, and:

$U_{2} = m\cdot g \cdot \Delta y$

Where:

$m$ - Mass of the bullet, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

$K_{1} = m\cdot g \cdot \Delta y$

$\Delta y = \frac{K_{1}}{m\cdot g}$

If $K_{1} = 1066\,J$ , $m = 0.25\,kg$ and $g = 9.81\,\frac{m}{s^{2}}$ , the maximum height is now computed:

$\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 434.791\,m$

$\Delta y = 0.435\,km$

Lastly, the range of the bullet is 0.435 kilometers.

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