Real world System of equations

Explain how surface waves can have characteristics of both longitudinal waves and transverse waves. Please use 3 content related

bonufazy [111]

Answer: Search Results

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Answer: Surface waves can have characteristics of both longitudinal and transverse waves in the following way; The motion of the surface waves is up and down which is perpendicular to the direction of the wave. This is similar to the motion of transverse waves whereas the the motion of longitudinal.

Explanation:

7 0

3 years ago

What is the wavelength of a wave that has a speed of 26 m/s and a frequency of 49? ? Show your work

nirvana33 [79]

Answer:

Wave speed = Wavelength x Frequency

26 m/s = Wavelength x 49

Divide by 49 to find the wavelength:

The wavelength is approximately 0.53

Let me know if this helps!

6 0

3 years ago

If the intensity of a loud car horn is 0.005 W/m^2 when you are 2 meters away from the source. Calculate the sound intensity lev

nekit [7.7K]

Answer:

(c) 97 dB sound intensity level

Explanation:

We have given the intensity of the loud car horn $I=0.005w/m^2$

We know that $I_O=10^{-12}w/m^2$

Now the sound intensity level is given by $\beta =10log\frac{I}{I_0}=10log\frac{0.005}{10^{-12}}=96.98dB$ , which is nearly equal to 97

So the sound intensity level will be 97 dB

So option (c) will be the correct option

4 0

3 years ago

Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy

BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

$T^{2} = a^{3}$

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.

4 0

3 years ago

An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi

elena55 [62]

Answer:

a) $\vec F_{B} = 8.766\times 10^{-14}\,T\,k$ , b) $\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

$\vec F_{B} = q\cdot \vec v \times \vec B$

The charge of the electron is equal to $-1.602\times 10^{-19}\,C$ . Then, cross product can be solved by using determinants:

$\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = 8.766\times 10^{-14}\,T\,k$

b) The charge of the proton is equal to $1.602\times 10^{-19}\,C$ . Then, cross product has the following determinant:

$\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

8 0

3 years ago