A 124 kg balloon carrying a 22 kg basket is descending with a constant downward velocity of 20 m/s. A 1 kg stone is thrown from the basket with an initial velocity of 15 m/s perpendicular to the path of descent. A person down below measures that it takes the stone 5 s to hit the ground.

How high is the balloon when the stone is thrown out of the basket?

Using the formula for displacement in one dimensional motion, we can calculate the height of the balloon to be 22.5 meters when the stone was thrown.

Explanation:

The subject of this question pertains to Physics, specifically the PHY-1.2.2 (Motion in One Dimension) topic.

To find how high the balloon was when the stone was thrown, we use the formula in motion which is h = ut + 0.5gt² where h is the height, u is the initial velocity, g is the acceleration due to gravity (about 9.8 m/s²) and t is the time taken.

The initial velocity u is given by the descent velocity of the balloon which is 20 m/s (negative since it is downwards), g=9.8 m/s² (positive since the balloon is descending) and t=5 s.

Substituting the values: h = (-20 m/s * 5 s) + 0.5 * 9.8 m/s² * (5 s)² = -100 m + 122.5 m = 22.5 m

So, the basket was thrown from a height of 22.5 m above the person who is on ground.

On the other hand, as you go across a row from left to right, the atomic radii decrease. This is because the valence electrons are in the same shell while more protons are added to the nucleus, increasing the pull on the valence electrons.

Overall, the atomic size generally increases from top to bottom in a group and decreases from left to right in a row on the periodic table.

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