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4 years ago

Helpp me pleassee....

Physics

1 answer:

lilavasa [31]4 years ago

4 0

Answer:

The fundamental wavelength of the vibrating string is 1.7 m.

Explanation:

We have,

Velocity of wave on a guitar string is 344 m/s

Length of the guitar string is 85 cm or 0.85 m

It is required to find the fundamental wavelength of the vibrating string. The fundamental frequency on the string is given by :

$f=\dfrac{v}{2l}\\\\f=\dfrac{344}{2\times 0.85}\\\\f=202.35\ Hz$

Now fundamental wavelength is :

$\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{344}{202.35}\\\\\lambda=1.7\ m$

So, the fundamental wavelength of the vibrating string is 1.7 m.

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3 years ago

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Answer:

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3 years ago

If the resultant of two velocity vectors of equal magnitude is also of the same magnitude, then which statement must be correct?

Tamiku [17]

The correct option is C) The angle between the vectors is 120°.

Why?

We can solve the problem and find the correct option using the Law of Cosine.

Let A and B, the given two sides and R the resultant (sum),

Then,

$R=A=B$

So, using the law of cosines, we have:

$R^{2}=A^{2}+B^{2}+2ABCos(\alpha)\\ \\A^{2}=A^{2}+A^{2}+2*A*A*Cos(\alpha)\\\\0=A^{2}+2*A^{2}*Cos(\alpha)\\\\Cos(\alpha)=-\frac{A^{2}}{2*A^{2}}=-\frac{1}{2}\\\\\alpha =Cos(-\frac{1}{2})^{-1}=120\°$

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Have a nice day!

4 0

3 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$