Kepler noticed an imaginary line drawn from a planet to the Sun and this line swept out an equal area of space in equal times, If we then draw a triangle out from the Sun to a planet’s position at one point in time, it is notice that the area doesn't change even after the planet has left the original position say like after 2 to 3days or 2hours. So to have same area of triangle means that the the planet move faster when that are closer to the sun and slowly when they are far from the sun.
This led to Kepler's law of orbital motion.
First Law: Planetary orbits are elliptical with the sun at a focus.
Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times.
Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semi-major axis is the same for all planets.
It is this Kepler's law that makes Newton to come up with his own laws on how planet moves the way they do.
Explanation:
Given that,
(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :
(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :
Therefore, this is the required solution.
The MA is 6! Hope This Helps!
===> Distance fallen from rest in free fall =
(1/2) (acceleration) (time²)
(122.5 m) = (1/2) (9.8 m/s²) (time²)
Divide each side by (4.9 m/s²): (122.5 m / 4.9 m/s²) = time²
(122.5/4.9) s² = time²
Take the square root of each side: 5.0 seconds
===> (Accelerating at 9.8 m/s², he will be dropping at
(9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'. We'll need this number for the last part.)
===> With no air resistance, the horizontal component of velocity
doesn't change.
Horizontal distance = (10 m/s) x (5.0 s) = 50 meters .
===> Impact velocity = (10 m/s horizontally) + (49 m/s vertically)
= √(10² + 49²) = 50.01 m/s arctan(10/49)
= 50.01 m/s at 11.5° from straight down,
away from the base of the cliff.
