All categories

3 years ago

Which answer choice provides the best set of labels for Wave A and Wave B?

Physics

2 answers:

Blizzard [7]3 years ago

4 0

Answer:

Answer Choice D:

Wave A: Low Frequency Wave

Wave B: High Frequency Wave

Explanation:

Lower Frequency waves have a higher wavelength, while higher frequency waves have a lower wavelength.

hodyreva [135]3 years ago

3 0

A has less energy and lower frequency, while B has greater energy and higher frequency.

You might be interested in

Wind erosion can be reduced by _____.

yanalaym [24]

Using land according to its capability. protect the soil surface with some form of cover. control runoff before it develops into an erosive force

4 0

3 years ago

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

gayaneshka [121]

Answer:

The volume is decreasing at 160 cm³/min

Explanation:

Given;

Boyle's law, PV = C

where;

P is pressure of the gas

V is volume of the gas

C is constant

Differentiate this equation using product rule:

$V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}$

Given;

$\frac{dP}{dt}$ (increasing pressure rate of the gas) = 40 kPa/min

V (volume of the gas) = 600 cm³

P (pressure of the gas) = 150 kPa

Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( $\frac{dv}{dt}$ );

(600 x 40) + (150 x $\frac{dv}{dt}$ ) = 0

$\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min$

Therefore, the volume is decreasing at 160 cm³/min

3 0

2 years ago

A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$