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AnnZ [28]
4 years ago
13

What is the difference between sound waves and water waves

Physics
1 answer:
Iteru [2.4K]4 years ago
8 0
Sound waves are three dimensional they spread out in all directions from the source of the sound. 
and water waves are essentially two dimensional i.e., on the surface of water.
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What current is used to power the United States power grid?
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The answer is Alternating Current
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Suppose that when you ride on your 7.50 kg bike the weight of you and the bike is supported equally by the two tires. If the gau
natali 33 [55]

Answer: 567 N

Explanation:

If the weight of the person and the bike are supported equally by the two tires, this means that the force acting on each tire, is half of the total weight.

We know that the gauge pressure in one tire, by definition, is equal to the force on the tire, over the contact surface between each tire and the road, so we can write:

P= F/A = (mrider g + mbike g) / A = (Fgrider + 7.5 kg. 9.8 m/s2) / 703 mm2 (1)

As the pressure data is given in lb/in2, it is needed to convert to N/mm2, as follows:

65.5 lb/in2 = 0.455 N/mm2.

Replacing in (1), and solving for  Fgrider, we have:

Fgrider = 567 N

4 0
3 years ago
What is the net force on this object?
vaieri [72.5K]

Answer:

200 newtons

Explanation:

because the sub air that would pull the force down by all of the mass of the sub air go down by that 400 newtons there for your anwer is 200 newtons. because 600-400=200

8 0
3 years ago
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Two mugs of tea were made and their temperatures measured. Which statement correctly describes the characteristics of the partic
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Its b on ed-genuity- The particles in mug A are moving more slowly.

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4 years ago
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When light with a wavelength of 238 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic ener
erastova [34]

Answer:

The wavelength is 173 nm.

Explanation:

This kind of phenomenon is known as photoelectric effect, it occurs when photons of light inside the metal surface and if they have the right amount of energy electrons absorb it and got expelled from the metal as photo electrons. The maximum kinetic energy of that photo electrons is given by the expression:

K_{max} =E_{photon} - \Phi (1)

With E the energy of the photon and Φ the work function of the material. The work function is a value characteristic of each material and is related with how much the electron is attached to the material, the energy of the photon is the Planck's constant (h=6.63\times10^{-34}) times the frequency of light (\nu) , then (1) is:

K_{max} =h\nu - \Phi (2)

The frequency of an electromagnetic wave is related with the wavelength (\lambda) by:

\nu=\frac{c}{\lambda} (3)

with c the velocity of light (c=3.0\times10^{8})

Using (3) on (2):

K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \Phi:

\Phi=\frac{hc}{\lambda}-K_max=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{238\times10^{-9}}-3.13\times10^{-19}

\Phi=5.23\times10^{-19} J

That's the work function of the metal we're dealing. So now if we want to know the wavelength to obtain the double of the kinetic energy we use:

2K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \lambda:

\lambda = \frac{hc}{2K_{max}+\Phi}=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{2(3.13\times10^{-19})+5.23\times10^{-19}}=1.73\times10^{-7}

\lambda=173 nm

3 0
4 years ago
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