Displacement is B) the shortest distance between the starting point and the ending point of a motion
Explanation:
Displacement is a vector quantity; it is a vector connecting the initial position to the final position of motion of an object.
Since it is a vector, it has both a magnitude and a direction:
- The magnitude of the displacement is the length of the vector, therefore it corresponds to the shortest distance in a straight line between the starting point and the ending point of the motion
- The direction goes from the starting point to the ending point
Therefore, the correct answer is
B) the shortest distance between the starting point and the ending point of a motion
Note that displacement is very different from distance. Consider for example an object moving in a circle, returning to its initial position: in this case, the distance covered by the object is not zero (it is the length of the circle), however the displacement is zero, because the initial position corresponds to the ending position.
Learn more about distance and displacement:
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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)
I believe the correct answer is C
Variables:
Source charge, Q = 3 micro C = 3 * 10^ - 6 C
E = electric field = 2.86 * 10 ^5 N/C
K = 8.99 * 10^9 N * m^2 / C
d = distance = ?
Formula:
E = K * Q / (d^2) => d^2 = K * Q / E
=> d^2 = 8.99 * 10^9 N * m^2 / C * 3 * 10^ -6 C / (2.86 * 10^ 5 N/C)
d^2 = 9.43 * 10 ^ -2 m^2
=> d = 3.07 * 10^-1 m
Answer: 0.307 m
Note: it is a long distance due to the Electric field is very low
