Although they're all 'close', none of the planets orbits in the same plane as any other planet. They're all in slightly different planes.
The farthest out compared to all the others is Pluto, with an orbit inclined about 17 degrees compared to the ecliptic plane (Earth's orbit). But Pluto is officially not a planet, so I don't think it's a good answer.
The next greatest inclination compared to Earth's orbit is <em>Mercury</em>. That one is about 7 degrees.
The other six planets are all in different orbital planes inclined less than 7 degrees compared to Earth's orbit.
The (more) speed an object has, the (lower) the potential energy and the (higher) the kinetic energy. (I believe that is correct but it’s been a while since I’ve done this)
The increase in speed leads to an increase in the amount of air resistance. Eventually, the force of air resistance becomes large enough to balances the force of gravity. At this instant in time, the net force is 0 Newton; the object will stop accelerating. The object is said to have reached a terminal velocity.
The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.
<h3>What do you mean by electric potential? </h3>
The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.
V = kq/r
Where V represents electric potential, K is coulomb constant, q is Charge and r is distance between any two around charge to the point charge.
Electric potential at O due to four charges is given by,
V = 4KQ/ r
where, r = √2a/2 = a/√2
V = 4k × Q√2/a
V = √2Q/πÈa
The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.
To learn more about electric potential refer to:
brainly.com/question/12645463
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Answer:
As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.
(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.
(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.
(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.
(iv) At point A, the driver will feel the lightest.
(v)The car can go that much fast without losing contact with the road at A can be determined as follow:
Fn=0 - lose contact with road
Fg= mv²/r
mg=mv²/r
v=sqrt (gr)
