Answer:
a
b
Explanation:
From the question we are told that
Their distance apart is 
The wavelength of each source wave 
Let the distance from source A where the construct interference occurred be z
Generally the path difference for constructive interference is
Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0
so
=> 
=>
Generally the path difference for destructive interference is
=> 
=> 
substituting values
=> 
So
and
=> 
=>
The unmagnetized pieces of iron would be randomly pointing to directions, this is true because although influenced with the magnetic domain, the direction of the unmagnetized iron field of attraction is not uniform or does not have preferred direction.
Answer:
Sound intensity levels are quoted in decibels (dB) much more often than sound intensities in watts per meter squared. Decibels are the unit of choice in the scientific literature as well as in the popular media. The reasons for this choice of units are related to how we perceive sounds. How our ears perceive sound can be more accurately described by the logarithm of the intensity rather than directly to the intensity. The sound intensity level β in decibels of a sound having an intensity I in watts per meter squared is defined to be β(dB)=10log10(II0)β(dB)=10log10(II0), where I0 = 10−12 W/m2 is a reference intensity. In particular, I0 is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. Sound intensity level is not the same as intensity. Because β is defined in terms of a ratio, it is a unitless quantity telling you the level of the sound relative to a fixed standard (10−12 W/m2, in this case). The units of decibels (dB) are used to indicate this ratio is multiplied by 10 in its definition. The bel, upon which the decibel is based, is named for Alexander Graham Bell, the inventor of the telephone.
Table 1. Sound Intensity Levels and IntensitiesSound intensity level β (dB)Intensity I(W/m2)Example/effect01 × 10–12Threshold of hearing at 1000 Hz101 × 10–11Rustle of leaves201 × 10–10Whisper at 1 m distance301 × 10–9Quiet home401 × 10–8Average home501 × 10–7Average office, soft music601 × 10–6Normal conversation701 × 10–5Noisy office, busy traffic801 × 10–4Loud radio, classroom lecture901 × 10–3Inside a heavy truck; damage from prolonged exposure[1]1001 × 10–2Noisy factory, siren at 30 m; damage from 8 h per day exposure1101 × 10–1Damage from 30 min per day exposure1201Loud rock concert, pneumatic chipper at 2 m; threshold of pain1401 × 102Jet airplane at 30 m; severe pain, damage in seconds1601 × 104Bursting of eardrums
Answer:
F = 3.20 N
Explanation:
Given:
Work done by child = 80.2 j
Distance that the car moves = 25.0 m
We need to find the force acting on the car.
Solution:
Using work done formula as.
Where:
W = Work done by any object.
F = Force (push or pull)
d = distance that the object moves.
Substitute 
in work done formula.
F = 3.20 N
Therefore, force acting on the car F = 3.20 N
