Answer:

Explanation:
Given data
Electric potential at point a is Ua=5.4×10⁻⁸J
q₂ moves to point b where a negative work done on it
Required
Electric potential energy Ub
Solution
When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

Now substitute the given values
So

Answer:
Kepler's First Law: each planet's orbit about the Sun is an ellipse. The Sun's center is always located at one focus of the orbital ellipse. The Sun is at one focus. The planet follows the ellipse in its orbit, meaning that the planet to Sun distance is constantly changing as the planet goes around its orbit.
Complete Question:
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.
Answer:
m = 0.001 M
For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/
V = l x w x h
v = 3 x 6 x 4
v = 72