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matrenka [14]
3 years ago
14

The part of a sound wave in which the particles are most spread out is called a(n)

Physics
1 answer:
Nutka1998 [239]3 years ago
7 0
Seismic wave is the answer
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.physics wave. Please don't dare to answer irrelevant
MAVERICK [17]

Answer:

1a) 857143 m

1b) 414 m

2a)

2b)

3) the medium of air has a wavelength of 0.334 m, the medium of water has a wavelength of 1.493 m, and the medium of 5.130 m.

Explanation:

Question 1a)

Given the velocity/speed, and frequency of the wave, the length can be calculated using these two quantites.

[ λ = v / f ] wavelength = <u>v</u>elocity of the wave / <u>f</u>requency of the wave in Hz.

Since 3 × 10^8 × ms^-1 is the velocity, and 350Hz is the frequency.

Anything to the negative power is reciprocated. i.e ms^-1 = m/s.

The wavelength is 300000000m/350Hz = 857142.8571428..... m ≈ 857143 m

Question 1b) Given that the frequency of the second wave in water is 1% of the first wave, and the speed of the second wave is 1450ms^-1

Therefore the second wave has a frequency of 1% of 3.5 = 350/100 Hz = 3.5 Hz

The wavelength is found using the same

formula: wavelength = 1450m/3.5Hz = 414.2857142857.... m ≈ 414 m

Question 2a)

Question 2b)

Question 3) Remember, the speed of sound of the medium = frequency of the medium × wavelength of the medium.

Therefore the wavelength of the medium = speed of sound of the medium / frequency of the medium. This has a similar correlation to the wavelength formula. We are given that all these mediums have a frequency of 1KHz = 1000Hz, where So the wavelength of each medium =

Question 4)

3 0
3 years ago
Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a
Nat2105 [25]

Answer:

Explanation:

Given

diameter d=20 \mu m

density \rho =1300 kg/m^3

frequency \nu =150 Hz

Length of silk strand L=14 cm

Velocity in the string is as follows

\nu =\sqrt{\frac{T}{\mu }}

The expression for Fundamental Frequency

f=\frac{\nu }{2l}

f=\frac{1}{2l}\times \sqrt{\frac{T}{\mu }}

f=\frac{1}{2l}\times \sqrt{\frac{T}{\frac{m}{l}}}

f=\frac{1}{2l}\times \sqrt{\frac{Tl}{\rho V}}

Squaring

f^2=\frac{1}{4l^2}\times \frac{Tl}{\rho V}

T=4\rho \cdot A\cdot l^2\cdot f^2

T=4\times 1300\times \frac{\pi }{4}(20\times 10^{-6})^2\times (0.14)^2\times 150^2

T=7.2\times 10^{-4} N

8 0
3 years ago
A ball rolls for 8 seconds and travels 24 meters. How fast was it traveling?
belka [17]

Answer:

The speed of the ball was, v = 3 m/s

Explanation:

Given data,

The time period of the ball, t = 8 s

The distance the ball rolled, d = 24 m

The velocity of an object is defined as the object's displacement to the time taken. The formula for the velocity is,

                              v = d / t      m/s

Substituting the given values in the above equation,

                               v = 24 / 8

                                  = 3 m/s

Hence, the speed of the ball was, v = 3 m/s

8 0
3 years ago
A 1.3-kg model airplane flies in a circular path on the end of a 23-m line. The plane makes
storchak [24]

(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in

(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s

(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:

(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s

(c) The plane accelerates toward the center of the path with magnitude

<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²

(d) By Newton's second law, the tension in the line is

<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N

4 0
3 years ago
(2.437×10⁴)(6.5411 x 10^9)/(5.37x10^6). write in scientific notation​
Musya8 [376]

Answer:

the answee is

2.968456 ×10^7

4 0
3 years ago
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