Because of its Sharp tip the needle is able to put the force on a very small area of the cloth, producing a large pressure sufficient enough to pierce the cloth being stitched and allow to stitch the thread in and pull it out.
The frequency of oscillation on the frictionless floor is 28 Hz.
<h3>
Frequency of the simple harmonic motion</h3>
The frequency of the oscillation is calculated as follows;
f = (1/2π)(√k/m)
where;
- k is the spring constant
- m is mass of the block
f = (1/2π)(√7580/0.245)
f = 28 Hz
Thus, the frequency of oscillation on the frictionless floor is 28 Hz.
Learn more about frequency here: brainly.com/question/10728818
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Answer: Ok, first lest see out problem.
It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.
Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0
where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.
So Q= rho*volume= pi*r*r*L*rho
so replacing : E = (1/2)*r*rho/ε0
you may ask, ¿why dont use R on the solution?
since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.
R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.
