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kumpel [21]
3 years ago
12

A 60cm long string of an ordinary guitar is tuned to produce the note a4 (frequency 440hz) when vibrating in its fundamental mod

e. if the speed of sound in the surrounding air is 344 m/s, find the wavelength of the sound wave produced in the air by the vibration of the string
0.99m
0.56m
0.78m
1.12m
Physics
2 answers:
pishuonlain [190]3 years ago
8 0

Answer:

0.78 m

Explanation:

The relationship between wavelength and frequency of a wave is given by

v=f \lambda

where

v is the speed of the wave

f is the frequency

\lambda is the wavelength

For the sound wave in this problem, we have

f=440 Hz is the frequency

v = 344 m/s is the speed of sound in air

Substituting into the equation and re-arranging it, we find the wavelength:

\lambda=\frac{v}{f}=\frac{344 m/s}{440 Hz}=0.78 m

Mariana [72]3 years ago
8 0

Answer: 1. B) 720 N

2. D) 2500 m

3. D) 408 Hz

4. C) 4m

5. A) 246 hz

6. D) 31st

7. B) the frequency of the sound is 100 Hz

8. C) 0.78 m

9. B) 16 Hz

10. B) 311 m/s

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The maximum tensile stress immediately adjacent to the hole

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Explanation:

To solve the question we have

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Width of steel bar = 4 in

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Available width at cross section where the 1.00 in diameter hole is drilled =

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Cross sectional area at the point of reduced cross section due to the drilled hole = Width × Thickness (Since the item is a flat bar)

= 3 in × 1/2 in = 1.5 in²

Therefore Stress = (30000 lb)/(1.5 in²) = 20000 lb/in²

the maximum tensile stress immediately adjacent to the hole.

Bending stress = \sigma_B= \frac{M_y}{I} where I = \frac{(0.5^2 + 4^2)}{12}

0.5*30000/I = 11076.92 lb/in²

Max stress = 31076.92 lb/in²

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3 years ago
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Answer:

Yes, there is such a way.

Explanation:

If currents flow in the same direction in two or more long parallel wires, there will be an attractive force between the wires. If the current flows in different directions, there will be a repulsive force between the wires. In this case, these three parallel wires, can be be made to carry current in the same direction, creating an attractive force between all three wires.

Note that it is not possible to have at the least one of them carry current in the opposite direction and still have an attractive current between them.

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3 years ago
A 0.20-f capacitor is desired. what area must the plates have if they are to be separated by a 3.2-mm air gap?
Igoryamba
Air gap means that the dielectric is air.

So <span>ε0 = </span><span>8.85 x 10^-12 [F/m].......................permitivity of free space

Lets use the equation

</span>C= ( ε0x A) / d

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And d the distance between the plates

d = <span>3.2-mm = 3.2 E-3 m

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chapter 2 linear motion problems a student launches an arrow upward with an unknown initial velocity. the arrow takes 2.3 second
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Answer:

V_{0}= 22.5\frac{m}{s} and Ymax=25.8m

Explanation:

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