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3 years ago

6

Which correctly ranks, from fastest to slowest, the time it will take one wavelength to pass a certain point?

2 answers:

Andreyy893 years ago

7 0

Yulia observed a sound wave moving through a solid medium at four different temperatures, and she recorded observations about each one in the chart.

Which correctly ranks, from fastest to slowest, the time it will take one wavelength to pass a certain point?

Wave X → Wave Y → Wave W → Wave Z

Wave Z → Wave W → Wave Y → Wave X

Wave X → Wave W → Wave Y → Wave Z

Wave Z → Wave Y → Wave W → Wave X

dsp733 years ago

3 0

No option to choose from? I'd say history graph

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What do we call the principal SI units that are used to derive all other SI units? Question 3 options:

Angelina_Jolie [31]

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3 years ago

A radio station broadcasts its music with waves at a frequency of 7.34 x 10²Hz. These radio waves travel at a speed of 3.00x 10%

Lorico [155]

Answer:

See below

Explanation:

<u>I will use 3 x 10^8 m/s for speed or wave</u>

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2 years ago

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago