I’d think the answer would be C. i’m just kinda guessing but my thought process is this (as simply as i can put it because physics is confusing):
so for example say you throw a ball across a flat surface. inertia is what keeps the ball rolling straight in a line, so unless you were to maybe put your hand in front of the ball or something, it would just go straight forever.
this is what happens with the planets. they go in a straight line, but since there’s gravity, the planets are also being pulled towards the sun. so gravity and inertia are why the planets orbit in the circle pattern they do. so when we remove inertia, we’re removing the state in which the planets keep going straight while being pulled towards a center point (the sun). this causes gravity to be the only factor in the planets orbiting. so that being said, the planets would just be pulled towards the sun. :)
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Answer:

cubic metre or 1e-9
Explanation:
•By division. Number of cubic millimetre divided(/) by 1000000000, equal(=): Number of cubic metre.
•By multiplication. 83 mm3(s) * 1.0E-9 = 8.3E-8 m3(s)
Answer:
vf = 3.27[m/s]
Explanation:
In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.
With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.
With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.
Then we equalize the two stress equations and we can clear the acceleration.
a = 3.58 [m/s^2]
As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.
![x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2Aa%2At%5E%7B2%7D%5C%5C1.5%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A%283.58%29%20%2At%5E%7B2%7D%20%5C%5Ct%20%3D%200.91%20%5Bs%5D)
And the speed can be calculated as follows:
![v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Ba%2At%5C%5Cv_%7Bf%7D%3D0%2B%283.58%2A0.915%29%5C%5Cv_%7Bf%7D%3D%203.27%5Bm%2Fs%5D)