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3 years ago

1 AI A3 If A1 = 27A and A3 = 1.9A, what would A2 read?

Physics

1 answer:

Gnoma [55]3 years ago

7 0

Answer:

0.8A

Explanation:

because im very intelligent and beautiful so the answer is correct and its 0.8A

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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 oscill

avanturin [10]

Answer:

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 $7850 kg/m^3$

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

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

$w$ is velocity of the wave

This is mathematically evaluated as

$w=2 \pi f$

substituting 60Hz for f

We have

$w = 2 *3.142 * 60$

$=377.04 \ rad/s$

$A$ 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 $\frac{1}{2}$

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