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Genrish500 [490]
4 years ago
13

james wants to lower the pitch coming from an electronic sound generator. how would he achieve that? a. he should increase the a

mplitude. b. he should increase the frequency c. he should decrease the amplitude. d. he should decrease the frequency. e. he should increase the intensity.
Physics
2 answers:
Gnesinka [82]4 years ago
3 0

Answer: Option D:<u> He should decrease the frequency. </u>

Pitch is a perceptual property of sound. It can be perceived in the sense of high or low musical melodies. The pitch and frequency are related. when frequency increases, pitch increases. And when frequency decrease, pitch decreases.

Hence, the correct option is D: he should decrease the frequency.

xxTIMURxx [149]4 years ago
3 0
He should decrease the frequency because frequency is the pitch of a sound so the lower the frequency the lower the pitch
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The Milky Way is changing because a large numbers of stars are undergoing blue-shift.

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Two metra trains approach each other on separate but parallel tracks. one has a speed of 90 km/hr, the other 80 km/hr. initially
Gennadij [26K]

The trains take <u>57.4 s</u> to pass each other.

Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.

therefore,

v_A=90 km/h\\ v_B=-80 km/h

The relative velocity of the train A with respect to B is given by,

v_A_B=v_A-v_B\\ =(90km/h)-(-80km/h)\\ =170km/h

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.

The trains are a distance d = 2.71 km apart.

Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

t= \frac{d}{v_A_B}

Substitute 2.71 km for d and 170 km/h for v_A_B

t= \frac{d}{v_A_B}\\ =\frac{2.71 km}{170 km/h} \\ =0.01594 h

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6 0
3 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
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Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
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