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diamong [38]
4 years ago
10

5 A \machine produces compression waves in a spring that is 120 cm long by pulsing twice every second. The back and forth moveme

nt of the pulse creates compressed sections of the spring that are 30 cm apart and travel toward the other end of the spring. Which property of the wave does the value of two per second describe? A. wavelength B. intensity C. frequency D. amplitude
Physics
2 answers:
Lisa [10]4 years ago
8 0

Answer:

D. Frequency

Explanation:

The machine produces compression waves in a spring that is 120cm long and it pulse twice every seconds. The property the wave exhibits by exhibiting the value of two per second is the Frequency.

Amplitude of a a wave is the maximum displacement from the equilibrium position. The amplitude of an object is equal to one half the path of the vibration path. The amplitude of a wave is proportional to the amplitude of the source. Amplitude measures how big a wave is . Amplitude of a wave usually look at the energy of a wave so it is not the amplitude .

The frequency of a wave gives the number of  wave that passed fixed point at a given time.

olga2289 [7]4 years ago
7 0
The correct answer is C) frequency.
In fact, the frequency is the number of wave crests (or pulses) per seconds. In our problem, the machine that produces the wave pulses two times per second, so this is exactly the frequency of the compression wave.
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A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant
OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

a_t = ar

where;

a is the angular acceleration and

r is the radius of the circular path

a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2

Determine time of the rotation;

\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2

The magnitude of total linear acceleration is given by;

a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27  \ m/s^2

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

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