Physics, 08.04.2020 03:26 chrissycherry

A small steel wire of diameter 1.0 mm is connected to an oscillator and is under a tension of 7.5 N. The frequency of the oscillator is 60.0 Hz and it is observed that the amplitude of the wave on the steel wire is 0.50 cma. What is the power output of the oscillator, assuming that the wave is not reflected back?b. If the power output stays constant but the frequency is doubled, what is the amplitude of the wave?

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Answer from: carnations

The output power is P= 0.764Watt

The Amplitude would decrease by $\frac{1}{2}$

Explanation:

From the question we are told that

The diameter of the steel wire is = 1.0mm $= \frac{1}{1000} = 1.0*10^{-3}m$

The raduis of this steel wire is $r = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Now from the question we can deduce that the power output is equal to the power being transmitted by wave on the wire this is mathematically represented as

$P = \frac{1}{2} \mu w^2 A^2 v ----(1)$

Where $\mu$ is the mass per unit length of the wire

This is mathematically evaluated as

$\mu = a* \rho$

Where a is the area of the the wire = $\pi r^2 = (3.142 * 0.5*10^{-3})^2 =7.855*10^{-7}m^2$

$\rho$ is the density of steel with a generally value of 7850 kg/m^3

$\mu = 7.855*10^{-7} *7850$

$= 6.162*10^{-3}kg/m$

is velocity of the wave

This is mathematically evaluated as

$w=2 \pi f$

substituting 60Hz for f

We have

w = 2 *3.142 * 60

$=377.04 \ rad/s$

is the amplitude with a given value of 0.50 cm $= \frac{0.50}{100} = 0.50 *10^{-2}m$

v is the linear velocity of the wave

This is mathematically evaluated as

$v = \sqrt{\frac{T}{\mu} }$

Where T is the tension with a given value of 7.5N

$v = \sqrt{\frac{7.5}{6.162*10^{-3}} }$

=34.89 m/s

Substituting values into equation 1

$P = 6.162*10^{-3}* 377.04^2 * (0.5*10^{-2})^2 * 34.89$

P= 0.764Watt

Since the doubling of the frequency does not affect the amplitude and from equation one the output power is $\ \frac{1}{2}$ of the Amplitude, Then the Amplitude would decrease by $\frac{1}{2}$

Answer from: Quest

d) physical means can be used to separate the components of air.

Answer from: Quest

answer;

d. are made mostly of liquid

explanation;

-the general composition and structure planets includes; they have a thick gas and liquid layer covering a small,solid core. mostly the gases are hydrogen and helium atmosphere and layers of liquid hydrogen. for instance neptune and uranus are examples of the outer planets, they have a core covered by ice and ammonia.

-neptune's atmosphere is hydrogen and helium with a bit of methane so there is no oxygen to breath and the interior is made up of frozen water and ammonia with a rock and iron core. thus, there is no liquid water or solid surface to survive.

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