Does the planet exert a torque on the meteoroid with respect to the center of mass of the planet? Why or why not?

A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the fri

omeli [17]

Answer:

4500 N

Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

a = v^2/r

where v is the velocity of the body and r is the radius of the circumference:

Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

fs = μN = μmg

The car will not slide if f = fs, i.e.

fs = μmg = m v^2/r

That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration

fs = (1000 kg) * (30m/s)^2 / (200 m) = 4500 N

7 0

3 years ago

Read 2 more answers

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0

2 years ago

Name the property illustrated be each statement. If t - 13 = 52, then 52 = t - 13

Natali5045456 [20]

The property illustrated to each statement is commutative because it shows that for some value of t both statements are equal.

3 0

2 years ago

Read 2 more answers

A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons

Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

$K.E\geq P.E$

or

$\dfrac{1}{2}mv^2 \geq G\dfrac{Mm}{R}$

where $m$ is the mass of the asteroid, $R= 24,000,000\:m$ is its distance form earth's center, $M = 5.9*10^{24}kg$ is the mass of the earth, and $G = 6.7*10^{-11}m^3/kg\: s^2$ is the gravitational constant.