Answer:
The correct answer is Dean has a period greater than San
Explanation:
Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.
T² = (4π² / G M) r³
When applying this equation to our case, the planet with a greater orbit must have a greater period.
Consequently Dean must have a period greater than San which has the smallest orbit
The correct answer is Dean has a period greater than San
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Answer:
Electric field, E = 40608.75 N/C
Explanation:
It is given that,
Mass of electrons, 
Initial speed of electron, u = 0
Final speed of electrons, 
Distance traveled, s = 6.3 cm = 0.063 m
Firstly, we will find the acceleration of the electron using third equation of motion as :



Now we will find the electric field required in the tube as :



E = 40608.75 N/C
So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.
Frictional forces act in the direction opposite to the MOTION. That direction could be the same OR opposite to applied force.
-- If you push a loaded heavy wagon from behind, trying to get it going faster, friction is acting against you, opposite to your force.
-- If you push a loaded rolling heavy wagon from in front, trying to make it slow down, friction is acting with you, in the same direction as your force.
-- Opposite to the motion both times.