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saul85 [17]
3 years ago
6

______ interference occurs when two waves overlap and the resulting wave has a larger amplitude.

Physics
2 answers:
ratelena [41]3 years ago
8 0
Constructive is correct
9966 [12]3 years ago
4 0

Answer: The correct answer is constructive interference.

Explanation:

Interference is the phenomenon in which there is superposition of the two waves. Interference are of two types: Constructive interference and Destructive interference.

Constructive interference occurs when the there is a superposition of the two waves. The resulting wave has a large amplitude.

Destructive interference occurs when the there is a superposition of the two waves. The resulting wave has a low amplitude.

Therefore, Constructive interference occurs when two waves overlap and the resulting wave has a larger amplitude.

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2 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

f = f_0 (\frac{v_0}{v_0-v})

Here,

f_0 = Frequency of Source

v_s = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v  = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

f = f_0 (\frac{v_0}{v_0-v})

300 = f_0 (\frac{343}{343-v})

(300*343) - 300v = 343f_0

Now the second expression will be,

f' = f_0 (\frac{v_0}{v_0-v/2})

290 = (343)(\frac{v_0}{343-v/2})

290*343-145v = 343f_0

Dividing the two expression we have,

\frac{(300*343) - 300v}{290*343-145v} = 1

Solving for v, we have,

v = 22.12m/s

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0
3 years ago
A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.07 Hz. On this tray is an empty cu
Anettt [7]

Answer:

Explanation:

Given

Frequency of SHM is f=2.07\ Hz

Amplitude of SHM is A=3.13\ cm

Cup begins to slip when it overcomes the friction force

Friction force F_s=\mu mg

Applied force F=ma

ma=\mu mg

a=\mu g

and maximum acceleration during SHM is

a=A\omega ^2

a=A(2\pi f)^2

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a=5.296\ m/s^2

\mu =\frac{a}{g}

\mu =\frac{5.296}{9.8}=0.54

6 0
3 years ago
Read 2 more answers
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