Interference and diffraction are the phenomena that support only the wave theory of light. Options 2 and 3 are correct.
<h3 /><h3>What is the interference of waves?</h3>
The result of two or more wave trains flowing in opposite directions on a crossing or coinciding pathways. This phenomenon is known as the interference of waves.
The phenomenon of interference occurs when two wave pulses are traveling along a string toward each other.
The light wave hypothesis states that light behaves like a wave. Since light is an electromagnetic wave, it may be transmitted without a physical medium.
Light has magnetic and electric fields, much like electromagnetic waves do.
Transverse waves, such as those seen in light waves, oscillate in the same direction as the wave's path. A wave of light may experience interference as well as diffraction as a result of these properties.
All of the remaining options are the light phenomenon.
Hence, options 2 and 3 are correct.
To learn more about the interference of waves refer to the link;
brainly.com/question/16098226
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Answer:</h2>
1000th multiple of the standard reference level for intensities.
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Explanation:</h2>
The sound intensity level (β), measured in decibels, of a sound with an intensity of I is defined as follows;
β = 10 log (I / I₀) --------------------(i)
Where;
I₀ = reference intensity
Given from the question;
β = sound level = 30dB
Substitute this value into equation (i) as follows;
30 = 10 log (I / I₀)
Divide both sides by 3;
3 = log (I / I₀)
Take antilog of both sides;
10^(3) = (I / I₀)
1000 = I / I₀
Solve for I;
I = 1000I₀
Therefore the intensity of the sound is 1000 times the standard reference level for intensities (I₀)
<span>A. Copper is a good conductor of electricity because its atoms have a loosely held electron in their outer shell that is able to move freely to other atoms.</span>
Explanation:
12) q = mCΔT
125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)
T = 82.0°C
13) Solving for ΔT:
ΔT = q / (mC)
a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C
b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C
c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C
d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C
e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C
f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C
14) q = mCΔT
q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)
q = 502,000 J
20) q = mCΔT
q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)
q = 742,000 J
24) q = mCΔT
q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)
q = -0.091 J
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
