Question

11 multiple choice 2 points if you were able to go to the top of a tall building and simultaneously drop an apple and a basketball, they would both hit the gound at about the same instant because the force of gravity on them is equal. they have the same mass. air resistance on both is the same. they have the same amount of kinetic energy. 12 multiple choice 2 points the moon has less gravity than earth because the moon is farther from the sun. the moon has no atmosphere. the moon has less mass than earth. the moon is made of volcanic rock. 13 multiple choice 2 points why must the space shuttle use great force to get into outer space? earths spin pushes the shuttle. weather systems move near the shuttles launch area. earths gravity pulls on the shuttle. the shuttle is exposed to high temperatures.

Explanation:

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• The force of gravity \(F = mg\) (where \(m\) is mass and \(g\) is acceleration due to gravity). Since \(m\) (mass of apple and basketball) is different, \(F\) (force of gravity) is different.
• Mass of apple and basketball is different.
• Kinetic energy \(KE=\frac{1}{2}mv^{2}\). Since \(m\) (mass) is different and we are just dropping (initial \(v = 0\), but at impact \(v\) is same due to \(v=\sqrt{2gh}\), but \(m\) is different so \(KE\) is different).
• When we ignore air - resistance (in a simple free - fall model), the acceleration of an object in free - fall is \(a = g=\frac{F}{m}=\frac{mg}{m}\) (from \(F = mg\) and \(F=ma\)). So, the acceleration is independent of mass.

• The gravitational force \(F=\frac{GMm}{r^{2}}\) (for an object of mass \(m\) on the surface of a planet of mass \(M\) and radius \(r\)). The acceleration due to gravity \(g=\frac{GM}{r^{2}}\) (where \(G\) is the gravitational constant). Since the Moon has less mass \(M\) than the Earth ( \(r\) is also different but mass is the main factor here as \(g\propto M\) (when \(r\) is considered in a relative sense, but mass difference is more significant. The distance from the Sun (\(r_{sun - object}\)) does not affect the surface gravity of the Moon (surface gravity is \(g=\frac{GM_{moon}}{r_{moon}^{2}}\)).
• Having no atmosphere has no relation to surface gravity (\(g=\frac{GM}{r^{2}}\)).
• The composition (volcanic rock) has no relation to \(g=\frac{GM}{r^{2}}\).

• Earth's spin: The Earth's spin (angular motion) does not push the shuttle in the upward direction (in fact, for east - ward launches, we use the Earth's spin to get a boost in tangential velocity, but it's not a 'push' against gravity).
• Weather systems: Weather systems (like winds etc.) are negligible compared to the force needed to overcome gravity.
• High temperatures: High temperatures (due to air - friction during re - entry, but during launch, the main force is to overcome \(F = mg\) (weight). The gravitational force \(F = mg\) (where \(m\) is the mass of the shuttle) pulls the shuttle towards the Earth. To move into outer space (away from the Earth's gravitational pull), a force greater than \(mg\) (weight) is required.

Answer:

None of the options are correct in a strict sense (if we assume no air - resistance, the real reason is that the acceleration due to gravity is the same for all objects near the Earth's surface and \(h=\frac{1}{2}gt^{2}\) (for free - fall from rest, \(h\) is height, \(t\) is time). If we assume the problem is ignoring air - resistance, the closest is that the acceleration due to gravity is the same (but this option is not given. If forced to choose from given options, maybe a wrong - assumption - based view: if we wrongly assume \(F = mg\) and \(a=\frac{F}{m}\) and think about the equality of \(a\) in a wrong way related to the options, but actually, in reality, when we consider \(F = mg\) and \(F = ma\), \(a = g\) for all objects. But among the given options, if we assume a wrong - reasoning (like in some basic - level teaching where they just say 'g is same' and relate it to force in a wrong way), there is no correct option. But if we consider the formula \(F = ma\) and \(F = mg\), \(a = g\) (acceleration is same). But if we have to pick from the given:

• The force of gravity \(F=mg\) (if \(m\) is different \(F\) is different).
• Mass is different.
• Air - resistance is not same (different surface areas etc. but if we assume no air - resistance, this option is wrong).
• Kinetic energy \(KE=\frac{1}{2}mv^{2}\) ( \(m\) is different). So, there is an error in the problem's options. But if we consider the concept of acceleration due to gravity being same (derived from \(F = ma\) and \(F = mg\)), but no such option. If we assume the problem is in a very basic level (maybe a mis - question), and we know that \(a=\frac{F}{m}\) and \(F = mg\) gives \(a = g\) (acceleration same), but no option. If we consider the problem's options as wrong and the intended answer (maybe a mis - write) is related to the equality of acceleration (but not given). If we have to pick from the given, none is correct. But if we assume the problem is in a very basic course (like 5th grade) where they say 'g is same' and relate it wrongly to force (but \(F=mg\), \(m\) is different). So, actually, the problem has no correct option. But if forced (maybe a printing error), and assume the intended answer is related to the equality of acceleration (derived from \(F = ma\) and \(F=mg\)), but no option.

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