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Verizon [17]
4 years ago
7

The wavelength of a wave can be found by measuring how high the wave is measuring how many waves pass a point in a second measur

ing the distance from the crest of one wave to the crest of the next wave measuring the distance from the crest of one wave to the trough of the next wave
Chemistry
1 answer:
Liono4ka [1.6K]4 years ago
6 0

Now that you know something about the properties of the two main types of waves (Lesson 43), we need to make sure that you can look at individual characteristics that waves can have.

<span>Not all waves are created equal!<span>You need to be able to see the specific “faces” that each wave can have, based on three important characteristics: frequency, wavelength, and amplitude.</span></span>Frequency

When we first started looking at SHM we defined period as the amount of time it takes for one cycle to complete... seconds per cycle

<span><span>Frequency is the same sort of idea, except we’re just going to flip things around.</span><span>Frequency is a measurement of how many cycles can happen in a certain amount of time… cycles per second.</span><span>If a motor is running so that it completes 50 revolutions in one second, I would say that it has a frequency of 50 Hertz.</span><span>Hertz is the unit of frequency, and just means how many cycles per second.<span><span>It is abbreviated as Hz.</span><span>It is named after Heinrich Hertz, one member of the Hertz family that made many important contributions to physics.</span></span></span><span>In formulas frequency appears as an "f".</span></span>

Since frequency and period are exact inverses of each other, there is a very basic pair of formulas you can use to calculate one if you know the other…

It is very easy to do these calculations on calculators using the x-1 button.

Example 1: The period of a pendulum is 4.5s. Determine the frequency of this pendulum.

<span>The period means that it will take 4.5 seconds for the pendulum to swing back and forth once. So, I expect that my frequency will be a decimal, since it will complete a fraction of a swing per second.</span>Wavelength

Wavelength is a property of a wave that most people (once they know what to look for) can spot quickly and easily, and use it as a way of telling waves apart. Look at the following diagram...

Figure 1<span><span>Any of the parts of the wave that are pointing up like mountains are called crests. Any part that is sloping down like a valley is a trough.</span><span>Wavelength is defined as the distance from a particular height on the wave to the next spot on the wave where it is at the same height and going in the same direction.Usually it is measured in metres, just like any length.</span><span>There isn’t a special spot you have to start on a wave to measure wavelength, just make sure you are back to the same height going in the same direction. Most people do like to measure from one crest to the next crest (or trough to trough), just because they are easy to spot.</span></span>Figure 2

On a longitudinal wave, the wavelength is measured as the distance between the middles of two compressions, or the middles of two expansions.

Figure 3

This leads us to one of the most important formulas you will use when studying waves.

<span><span>Frequency tells us how many waves are passing a point per second, the inverse of time.</span><span>Wavelength tells us the length of those waves in metres, almost like a displacement.</span><span>If we multiply these two together, we are really multiplying 1/s and m… which gives us m/s, the velocity of the wave!</span></span>

v = velocity of the wave (m/s)
f = frequency (Hz)
λ = wavelength (m)

Example 2: A wave is measured to have a frequency of 60Hz. If its wavelength is 24cm, determine how fast it is moving.

Example 3: The speed of light is always 3.00e8 m/s. Determine the frequency of red light which has a wavelength of 700nm.

Be careful when changing the 700nm into metres. Some people get really caught up with changing it into regular scientific notation with only one digit before the decimal. Why bother? It's only being used in a calculation. You’ll probably just make a mistake changing the power of 10, so just substitute in the power for the prefix and leave everything else alone…700 nm = 700 x 10-9 m since “nano” is 10-9.

Amplitude

Amplitude is a measure of how big the wave is.

<span>Imagine a wave in the ocean. It could be a little ripple or a giant tsunami.<span>What you are actually seeing are waves with different amplitudes.<span>They might have the exact same frequency and wavelength, but the amplitudes of the waves can be very different.</span></span></span>

The amplitude of a wave is measured as:

<span><span>the height from the equilibrium point to the highest point of a crest or</span><span>the depth from the equilibrium point to the lowest point of a trough</span></span>Figure 4

When you measure the amplitude of a wave, you are really looking at the energy of the wave.

<span>It takes more energy to make a bigger amplitude wave.<span>Anytime you need to remember this, just think of a home stereo’s amplifier… it makes the amplitude of the waves bigger by using more electrical energy.</span></span>
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The following information is to be used for the next 2 questions. In order to analyze for Mg and Ca, a 24-hour urine sample was
Ainat [17]

Answer:

Explanation:

From the given information:

The concentration of metal ions are:

[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}

[Ca^{2+}]=0.007118 \ M

[Mg^2+] = \dfrac{0.003474 \ M\times (26.23  - 20.49 )mL}{10.0 \ mL}

=0.001994 \ M

Mass of Ca²⁺ in 2.00 L urine sample is:

= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}

= 0.1598 g

Mass of Ca²⁺ = 159.0 mg

Mass of Mg²⁺ in 2.00 L urine sample is:

= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

5 0
3 years ago
Calculate the percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl on the assumption that t
stealth61 [152]

Answer:

1.089%

Explanation:

From;

ν =1/2πc(k/meff)^1/2

Where;

ν = wave number

meff = reduced mass or effective mass

k = force constant

c= speed of light

Let

ν =1/2πc (k/meff)^1/2  vibrational wave number for 23Na35 Cl

ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl

The between the two is obtained from;

ν' - ν /ν  = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2

Therefore;

ν' - ν /ν = [meff/m'eff]^1/2 - 1

Substituting values, we have;

ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2  -1

ν' - ν /ν = -0.01089

percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;

ν' - ν /ν * 100

|(-0.01089)|  × 100 = 1.089%

4 0
3 years ago
A piece of glass is immersed in a liquid. It proceeds to float on the liquid's surface. This shows that the density of the glass
tamaranim1 [39]

Answer: b) Less dense

Explanation:

Differences in density is one reason objects float or sink.

An object more dense than the fluid in which it is immersed will sink, while objects less dense than the fluid in which it is immersed will float to the surface.

But objects floats at constant level if the density is equal to the density of the fluid in which it is immersed; it neither rises nor sinks in the fluid in this case.

7 0
3 years ago
The name of a binary compound ends with _____.<br><br> A.-ine<br> B.-ary<br> C.-ed<br> D.-ide
inn [45]
Cant really explain, for say, but its D.-ide

gfd
gd
fg
fg
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ds
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df
gd
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7 0
3 years ago
Light shining on a strip of metal can dislodge electrons. Do you think this is more consistent with light being made up of waves
Naya [18.7K]

Answer:

The correct answer to the following question will be "Particles".

Explanation:

  • A particle seems to be a little component of something, it's little. When you're talking about a subatomic particle, that would be a structured user likely won't see because it's quite unbelievably thin, but it has a tiny mass as well as structural integrity. Such particles seem to be tinier than that of the particles or atoms.
  • Such that the light which shines on the bit of metal could dissipate electrons, the particles seem to be more compatible with the light.
5 0
3 years ago
Read 2 more answers
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