Imagine that asteroid a that has an escape velocity of 50 m/s. if asteroid b has twice the mass and twice the radius, it would have an escape velocity the escape velocity of asteroid a.

The same as the escape velocity of asteorid A (50m/s)

Explanation:

The escape velocity is described as follows:

where is the universal gravitational constant, is the mass of the asteroid and is the radius

and since the scape velocity is 50m/s:

Now, if the astroid B has twice mass and twice the radius, we have that tha mass is:

and the radius is:

inserting these values into the formula for escape velocity:

and we have found that , so the two asteroids have the same escape velocity.

We found that the expression for escape velocity remains the same as for asteroid A, this because both quantities (radius and mass) doubled, so it does not affect the equation.

The answer is

Asteroid B would have an escape velocity the same as the escape velocity of asteroid A

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .