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Yuki888 [10]
3 years ago
9

A string, stretched between two fixed posts, forms standing-wave resonances at 325 Hz and 390 Hz. What is the largest possible v

alue of its fundamental frequency?
Physics
1 answer:
Pavel [41]3 years ago
3 0

Answer:

65

Explanation:

The resonant frequencies for a fixed string is given by the formula  nv/(2L).  

Where n is the multiple .

v is speed in m/s .

The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn

fundamental frequency means n=1

i.e  fn+1 – fn = 390 -325

                      =  65

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So, the dot product of V1 and V2 is = - 5 (3) + 7 (1) + 2 (2) = -15 + 7 + 4 = -4

The magnitude of vector V2 is given by

= \sqrt{3^{2}+1^{2}+2^{2}}=3.74

So, the projection of V1 along V2 = - 4 / 3.74 = - 1.07 ft

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so we need to find the direction of v2

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Pure water has a pH of 7. Pure water _______. A. Is a neutral substance B. Could be either an acid or a base C. Is a base D. Is
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A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir
mario62 [17]

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a . 0.35cm

b.  11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm

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