Answer:
things move quickly through a liquid
Explanation:
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
Using sources that are peer reviewed
Answer:
Explanation:
In a standing wave function characterized for x between (0.a). on the off chance that the amplitude of the wave interchange from positive to negative at the interval. there probably been a node at , among 0 and a to such an extent that . The reasoning is right that the likelihood of discovering the particle at the node is 0 in light of the fact that by definition, the nodes of the wave are the place where the wave function falls and is equivalent to 0. Since the likelihood of discovering a particle at a position at time , is provided by , this implies that at the nodes of a standing wave,
So the reasoning that the likelihood of the particle being at is 0 is right.
However, to examine whether the particle can travel from a position to a position of . All together words, can the molecule be found on one or the other side of the node?
The appropriate response is yes.
Recall that in quantum mechanics. wave functions at most present with the likelihood of discovering a particle at a specific time inside a time frame. The wave function doesn't present with an old classical actual trajectory that a particle should follow to go in space: all things being equal, it simply yields chances of whether a particle can be found in a specific spot at a specific time. So the reasoning that a particle can't get from a position to a position of , is incorrect.
Potential difference across the entire cd player is 12 V