(a) Are the radiating electric field lines around a charged particle straight lines when the particle is stationary?
1 answer:
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Answer:
E = 31.329 N/C.
Explanation:
The differential electric field at the center of curvature of the arc is
<em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>
where is the radius of curvature.
Now
,
where is the charge per unit length, and it has the value
Thus, the electric field at the center of the curvature of the arc is:
Now, we find and
. To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:
and this is
radians.
Therefore,
evaluating the integral, and putting in the numerical values we get:
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.
And the speed can be calculated as follows:
