A string with linear density 0.500 g/m.
Tension 20.0 N.
The maximum speed 
The energy contained in a section of string 3.00 m long as a function of .
We are given following data for string with linear density held under tension :
μ = 0.5 
= 0.5 x 10⁻³ 
T = 20 N
If string is L = 3m long, total energy as a function of is given by:
E = 1/2 x μ x L x ω² x A²
= 1/2 x μ x L x 
= 7.5 x 10⁻⁴
So, The total energy as a function of = 7.5 x 10⁻⁴
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Answer:
(1) 42.94 m
(2) 
Explanation:
Let us first draw a figure, for the given question as below:
In the figure, we assume that the person starts walking from point A to travel 11 m exactly 
south of west to point B and from there, it walks 21 m exactly 
west of north to reach point C.
Let us first write the two displacements in the vector form:
Now, the vector sum of both these vectors will give us displacement vector from point A to point C.
Part (1):
the magnitude of the shortest displacement from the starting point A to point the final position C is given by:
Part (2):
As the vector AC is coordinates lie in the third quadrant of the cartesian vector plane whose angle with the west will be positive in the north direction.
The angle of the shortest line connecting the starting point and the final position measured north of west is given by:
The portion of the flux leaves the curved surface of the cylinder is 60%.
<h3 /><h3>What are electrons?</h3>
The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.
If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the total flux leaving the cylinder.
If 20% of the flux leave from one end, then another 20% will leave from another end.
So, the net flux through curved surface is
100 -20 -20 = 60%
Thus, the total flux leaves the curved surface of the cylinder is 60%
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Answer: The second option.When
the current flows up the wire, the magnetic field flows out on the left
side of the wire and in on the right side of the wire.Explanation:
The first figure that I copy here with is the figure corresponding to this question.
The thumb is pointing upward.
The rule is that the thumb aims to the direction of the flow of current and the other fingers give the field lines.
The second figure that I attach is a free image from internet and it shows the direction of both the current and the fiedl lines.
So, the conclusion is that
the current goes upward the wire and the field lines go out of the paper (screen) for the points to the left of the wire and in on the right side of the wire.
A.) more weight lighting weights
