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Olenka [21]
3 years ago
7

What is an example of a mechanical noise?

Physics
1 answer:
faltersainse [42]3 years ago
4 0
Resonance:
The resounding recurrence is the recurrence at which a bit of metal, plastic or whatever else swings/vibrates with minimal measure of vitality input. Think about a man on a play area swing. You realize that it requires next to no push to keep the individual swinging. The recurrence at which they swing forward and backward is their full recurrence. In the event that you endeavor to influence them to swing speedier or slower, it will take altogether more vitality.

Resonating Panels:
This kind of clamor is caused when the bass notes are an indistinguishable recurrence from the thunderous recurrence of a metal or plastic board. To stop or decrease the commotion related with this kind of issue, you can do two or three things.

Rattling:
This sort of commotion would be caused when 2 bits of metal, plastic, whatever... are sufficiently close to hammer into each other when they resound. This is most likely best illuminated by filling the hole between the two vibrating parts with silicone sealant or shut cell froth climate stripping. The climate stripping is a superior arrangement in places like behind the tag. On the off chance that you have a tag outline, you can get some truly thin climate stripping and put between the casing and the plate.
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The circumference of a sphere was measured to be
professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

\phi = 76cm

Error (dr) = 0.5cm

The radius then would be

\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm

And

\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr

PART A ) For the Surface Area we have that,

A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)

Maximum error:

dA = 4(\frac{38}{\pi})(0.5)

dA = \frac{76}{\pi}cm^2

The relative error is that between the value of the Area and the maximum error, therefore:

\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}

\frac{dA}{A} = 0.01315 = 1.31\%

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

V = \frac{4}{3} \pi r^3

V = \frac{4}{3} \pi (\frac{38}{\pi})^3

V = \frac{219488}{3\pi^2}

Therefore the Maximum Error would be,

\frac{dV}{dr} = \frac{4}{3} 3\pi r^2

dV = 2r^2 (2\pi dr)

dV = 4r^2 (\pi dr)

Replacing the value for the radius

dV = 4(\frac{38}{\pi})^2(0.5)

dV = \frac{2888}{\pi^2} cm^3

And the relative Error

\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }

\frac{dV}{V} = 0.03947

\frac{dV}{V} = 3.947\%

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