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3 years ago

13

If a string vibrates at the fundamental frequency of 528 Hz and also produces an overtime with a frequency of 1,056 Hz this over

tune is the

1 answer:

Romashka-Z-Leto [24]3 years ago

5 0

I have a hunch that you're not talking about overtime
OR overtune. I think you're going for overtone .

For a fundamental frequency of 528 Hz,
1,056 Hz is the second harmonic.

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A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax

Colt1911 [192]

Answer:

From the question we are told that

The length of the rod is $L_o$

The speed is v

The angle made by the rod is $\theta$

Generally the x-component of the rod's length is

$L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

$L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=> $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

Generally the resultant length of the rod as seen by the observer is mathematically represented as

$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=> $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=> $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

$L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=> $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation:

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3 years ago

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mylen [45]

Answer:

because

Explanation:

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3 years ago

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2 years ago

What are 2 ways the velocity of an object can be changed

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3 years ago

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A technician wraps wire around a tube of length 33 cm having a diameter of 8.1 cm. When the windings are evenly spread over the

inessss [21]

Explanation:

It is given that,

Length of the tube, l = 33 cm = 0.33 m

Diameter of the tube, d = 8.1 cm = 0.081 m

Number of turns, N = 590

(a) We need to find the self- inductance of the solenoid. The formula for the self inductance is given by :

$L=\dfrac{\mu_o N^2A}{l}$

A is the area of the solenoid

$L=\dfrac{4\pi \times 10^{-7}\times (590)^2\times \pi \times (0.0405)^2}{0.33}$

L = 0.00683 H

or

L = 6.83 mH

So, the self inductance of the solenoid is 6.83 mH.

(b) The current in this solenoid increases at the rate of 3 A/s. $\dfrac{dI}{dt}=3\ A/s$

EMF in the solenoid is given by :

$E=-L\dfrac{dI}{dt}$

$E=-6.83\times 3$

E = -20.49 volt

Hence, this is the required solution.

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2 years ago