Halla el mínimo común múltiplo de 8

a small asteroid of mass 125 kg is orbiting a planet that has a mass of 3.52x 10^13 what is the radial distance between the aste

asambeis [7]

Answer:

r = 2.031 x 10⁶ m = 2031 km

Explanation:

In order for the asteroid to orbit the planet, the centripetal force must be equal to the gravitational force between asteroid and planet:

Centripetal Force = Gravitational Force

mv²/r = GmM/r²

v² = GM/r

r = GM/v²

where,

r = radial distance = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Planet = 3.52 x 10¹³ kg

v = tangential speed = 0.034 m/s

Therefore,

r = (6.67 x 10⁻¹¹ N.m²/kg²)(3.52 x 10¹³ kg)/(0.034 m/s)²

7 0

3 years ago

A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k

rewona [7]

The kinetic energy of the small ball before the collision is

   KE = (1/2) (mass) (speed)²

   = (1/2) (2 kg) (1.5 m/s)

   =    (1 kg) (2.25 m²/s²)

   =    2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

   Energy is never created or destroyed.
   If it seems that some energy disappeared,
   it actually had to go somewhere.
   And if it seems like some energy magically appeared,
   it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision. The large ball
and the small ball will just have to share the same 2.25 joules.

3 0

3 years ago

Match the indices of refraction with the corresponding effects on the waves.

Yanka [14]

Answer:

B) waves speed up

C) waves bend away from the normal

Explanation:

The index of refraction of a material is the ratio between the speed of light in a vacuum and the speed of light in that medium:

$n=\frac{c}{v}$

where

c is the speed of light in a vacuum

v is the speed of light in the medium

We can re-arrange this equation as:

$v=\frac{c}{n}$

So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is

B) waves speed up

Moreover, when light enters a medium bends according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1 ,n_2$ are the index of refraction of the 1st and 2nd medium

$\theta_1,\theta_2$ are the angles made by the incident ray and refracted ray with the normal to the interface

We can rewrite the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

So we see that if the index of refraction of the second medium is lower ( $n_2$ ), then the ratio $\frac{n_1}{n_2}$ is larger than 1, so the angle of refraction is larger than the angle of incidence: