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den301095 [7]
2 years ago
9

.

Physics
1 answer:
sergij07 [2.7K]2 years ago
4 0

Based on the nature of constructive interference,

none of the wave intersections will produce constructive interference.

<h3>What is constructive interference?</h3>

Constructive interference is interference that occurs when two waves of the same frequency amplitude and wavelength travelling in the same direction are superposed with the resultant effect of reinforcement of both waves to produce a larger wave.

Constructive interference occurs when the path difference between two identical waves at a point is:

  • delta s = n × wavelength

where n = 0, 1, 2, ...

For the various intersections:

  • 1.77 cm crest intersecting with a 0.65 cm crest; n = 1.77/0.65 = 2.7
  • A 1.2 mm crest intersecting with a 3.9 mm trough is destructive.
  • A 4.55 N trough intersecting with a 1.59 N trough; n = 4.55/1.59 = 2.8
  • A 0.44 inch trough intersecting with a 0.72 inch crest is destructive.
  • A 7.42 mm trough intersecting with a 1.93 mm trough; n = 7.42/1.93 = 3.8

Therefore, none of the wave intersections will produce constructive interference

Learn more about constructive interference at: brainly.com/question/1040831

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A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H
Aleks04 [339]

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .

Explanation:

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3 0
3 years ago
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VLD [36.1K]

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8 0
2 years ago
a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how
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Answer:

208.33 W

141.26626 seconds

Explanation:

E = Energy = 6\times 10^6\ J

t = Time taken = 8 h

m = Mass = 2000 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height of platform = 1.5 m

Power is obtained when we divide energy by time

P=\frac{E}{t}\\\Rightarrow P=\frac{6\times 10^6}{8\times 60\times 60}\\\Rightarrow P=208.33\ W

The average useful power output of the person is 208.33 W

The energy in the next part would be the potential energy

The time taken would be

t=\frac{E}{P}\\\Rightarrow t=\frac{mgh}{208.33}\\\Rightarrow t=\frac{2000\times 9.81\times 1.5}{208.33}\\\Rightarrow t=141.26626\ s

The time taken to lift the load is 141.26626 seconds

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