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motikmotik
3 years ago
6

A machine that promises more energy output than input is??

Physics
1 answer:
Sphinxa [80]3 years ago
3 0
-- sometimes presented as a "perpetual motion machine" /// -- impossible to construct /// -- a scam /// -- a fraud /// -- often lubricated with snake oil
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A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s
sergey [27]

Answer:

Y = V / f      where Y equals wavelength

4 Y1 = V / f1       for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2   for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1      dividing equations

f2 = 2 f1  

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

8 0
3 years ago
The force on an object is F⃗ =−17j⃗ . For the vector v⃗ =2i⃗ +3j⃗ , find: (a) The component of F⃗ parallel to v⃗
Igoryamba

Answer:

(a) \vec F_{\parallel} = -\frac{102}{13}\,i-\frac{103}{13}\,j , (b) \vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j, (c) W = -51

Explanation:

The statement is incomplete:

The force on an object is \vec F = -17\,j. For the vector \vec v = 2\,i +3\,j. Find: (a) The component of \vec F parallel to \vec v, (b) The component of \vec F perpendicular to \vec v, and (c) The work W, done by force \vec F through displacement \vec v.

(a) The component of \vec F parallel to \vec v is determined by the following expression:

\vec F_{\parallel} = (\vec F \bullet \hat {v} )\cdot \hat{v}

Where \hat{v} is the unit vector of \vec v, which is determined by the following expression:

\hat{v} = \frac{\vec v}{\|\vec v \|}

Where \|\vec v\| is the norm of \vec v, whose value can be found by Pythagorean Theorem.

Then, if \vec F = -17\,j and \vec v = 2\,i +3\,j, then:

\|\vec v\| =\sqrt{2^{2}+3^{3}}

\|\vec v\|=\sqrt{13}

\hat{v} = \frac{1}{\sqrt{13}} \cdot(2\,i + 3\,j)

\hat{v} = \frac{2}{\sqrt{13}}\,i+ \frac{3}{\sqrt{13}}\,j

\vec F \bullet \hat{v} = (0)\cdot \left(\frac{2}{\sqrt{13}} \right)+(-17)\cdot \left(\frac{3}{\sqrt{13}} \right)

\vec F \bullet \hat{v} = -\frac{51}{\sqrt{13}}

\vec F_{\parallel} = \left(-\frac{51}{\sqrt{13}} \right)\cdot \left(\frac{2}{\sqrt{13}}\,i+\frac{3}{\sqrt{13}}\,j  \right)

\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j

(b) Parallel and perpendicular components are orthogonal to each other and the perpendicular component can be found by using the following vectorial subtraction:

\vec F_{\perp} = \vec F - \vec F_{\parallel}

Given that \vec F = -17\,j and \vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j, the component of \vec F perpendicular to \vec v is:

\vec F_{\perp} = -17\,j -\left(-\frac{102}{13}\,i-\frac{153}{13}\,j  \right)

\vec F_{\perp} = \frac{102}{13}\,i + \left(\frac{153}{13}-17 \right)\,j

\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j

(c) The work done by  \vec F through displacement \vec v is:

W = \vec F \bullet \vec v

W = (0)\cdot (2)+(-17)\cdot (3)

W = -51

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