<h2>
Answer: Earth's orbital path around the Sun</h2><h2>
</h2>
The <u>Ecliptic</u> refers to the orbit of the Earth around the Sun. Therefore, <u>for an observer on Earth it will be the apparent path of the Sun in the sky during the year, with respect to the "immobile background" of the other stars.</u>
<u />
It should be noted that the ecliptic plane (which is the same orbital plane of the Earth in its translation movement) is tilted with respect to the equator of the planet about 
approximately. This is due to the inclination of the Earth's axis.
Hence, the correct option is Earth's orbital path around the Sun.
Answer:
955.5N
Explanation:
The normal force is given by the difference between the centripetal force and gravity at the top of the loop:
mass m = 65kg
radius of the loop r = 4m
velocity v = ?
g = 9.8 m/s²
To find the centripetal force, you need to find the velocity of the car at the top of the loop.
Use energy conservation:
At the top of the hill:
At the top of the loop:
Setting both energies equal and canceling the mass m gives:
Solving for v:
Using v in the first equation:
Given: The mass of stone (m) = 0.5 kg
Raised from heights (h₁) = 1.0 m to (h₂) = 2.0 m
Acceleration due to gravity (g) = 9.8 m/s²
To find: The change in potential energy of the stone
Formula: The potential energy (P) = mgh
where, all alphabets are in their usual meanings.
Now, we shall calculate the change in potential energy of the stone
Δ P = P₂ - P₁ = mg (h₂ - h₁)
or, = 0.5 kg ×9.8 m/s² ×(2.0 m - 1.0 m)
or, = 4.9 J
Hence, the required change in the potential energy of the stone will be 4.9 J
12.5 times 14 and convert to meters its 1.75 meters per second
