Ask question

Ask question

All categories

3 years ago

10

A 2.00 × 103 kg car rounds a circular turn of radius 20.0 m. If the road is flat and the coefficient of static friction between

the tires and the road is 0.70, how fast can the car go without skidding?

1 answer:

julsineya [31]3 years ago

3 0

Answer:

11.7 m/s

Explanation:

In order for the car to remain in circular motion along the road and not to skid, the frictional force between the tires and the road must be equal to the centripetal force.

Therefore, we can write:

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

where:

m = 2000 kg is the mass of the car

$\mu=0.70$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration due to gravity

v is the speed of the car

r = 20.0 m is the radius of the turn

Substituting and solving for v, we find the speed of the car:

$v=\sqrt{\mu gr}=\sqrt{(0.70)(9.8)(20.0)}=11.7 m/s$

You might be interested in

Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face

Sunny_sXe [5.5K]

Answer:

$\mu = 1.645$

Explanation:

By Snell's law we know at the left surface

$\theta_i = 19^o$

$\theta_r = ?$

$\mu_1 = 1$

$\mu_2 = \mu$

now we have

$1 sin19 = \mu sin\theta_r$

$0.33 = \mu sin\theta_r$

now on the other surface we know that

angle of incidence = $\theta_r'$

$\theta_e = 90$

so again we have

$\mu sin\theta_r' = 1 sin90$

so we have

$\theta_r = sin^{-1}\frac{0.33}{\mu}$

$\theta_r' = sin^{-1}\frac{1}{\mu}$

also we know that

$\theta_r + \theta_r' = 49$

$sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49$

By solving above equation we have

$\mu = 1.645$

3 0

2 years ago

Read 2 more answers

Kiley went 5.7 km/h north and then went 5.8 km/h west. From start to finish, she went 8.1 km/h northwest. Which best describes h

meriva

Answer:

8.1km/h Northwest

Explanation:

The 8.1km/h Northwest gives the best description of her distance from start to finish. This distance can be represented in a right angle triangle , this is the hypotenuse which is the longest side of the triangle. If we add 5.7 and 5.8 this gives 11.5km/h compared to 8.1km/h which is a smaller distance and the best.

6 0

2 years ago

which of these is an example of a biotic factor that could affect the size of a population A-climate b-disease c-a natural disas

andreyandreev [35.5K]

Hey there!

Biotic factors are living organisms. This includes plants, animals, and bacteria. Climate, water, and natural disasters are abiotic factors, which means they are nonliving factors that affect the environment. Disease is caused by bacteria, which is a biotic factor.

Therefore, your answer is B) Disease.

Hope this helps!

4 0

3 years ago

Are the two expressions equvalent 7(8x+5) and 48x + 35

Evgesh-ka [11]

Yes they are equivalent because 7x5=35 and 8x x 5=48x

4 0

3 years ago

A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba

Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

The mass of the rubber ball is $m_r = 2 \ kg$

The initial speed of the rubber ball is $u = 3 \ m/s$

The final speed at which it bounces bank $v - 3 \ m/s$

The mass of the clay ball is $m_c = 2 \ kg$

The initial speed of the clay ball is $u = 3 \ m/s$

The final speed of the clay ball is $v = 0 \ m/s$

Generally Impulse is mathematically represented as

$I = \Delta p$

where $\Delta p$ is the change in the linear momentum so

$I = m(v-u)$

For the rubber is

$I_r = 2(-3 -3)$

$I_r = -12\ kg \cdot m/s$

=> $|I_r| = 12\ kg \cdot m/s$

For the clay ball

$I_c = 2(0-3)$

$I_c = -6 \ kg\cdot \ m/s$

=> $| I_c| = 6 \ kg\cdot \ m/s$

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

8 0

3 years ago