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VladimirAG [237]

3 years ago

15

What are the first three harmonics of a note produced on a 0.31 m long violin string if the waves on this string have a speed of

274.4 M/s

1 answer:

Gelneren [198K]3 years ago

5 0

Answer:

442 Hz, 882 Hz, 1330 Hz

Explanation:

Given data: length of the string= 0.31 m = 31 cm

speed of the waves = 274.4 M/s or 2740 cm/s

Wavelengths are

62 cm

31 cm and

62 / 3 = 20.67 cm

The frequencies associated with these wavelengths ( which are the 1st, 2nd and 3rd harmonics )

are

27440/ 62 = 442.6 = 442 Hz

27440/31 = 885.2 = 882 Hz

27440 / 20.67 = 1328 = 1330 Hz

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What is 3600Hz has in rpm?

Setler [38]

Answer:

Explanation:

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3 years ago

These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the neg

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Answer:

Following are the solution to the given question:

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Calculating the common potential:

$\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\$

Calculating the charge after redistribution:

$When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2$

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6 0

3 years ago

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

a

The orbital speed is $v= 2.6*10^{3} m/s$

b

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

3 years ago