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klemol [59]
3 years ago
13

As ocean waves approach shore, their velocity decreases. How does a decrease in velocity affect the frequency and wavelength of

the waves entering the shallow water?
Physics
2 answers:
Elina [12.6K]3 years ago
6 0
Because if the velocity decreases then the frequency and wavelength would be affected since frequency deals with how fast the wave is going and since shallow water is lower theres not many waves to continue the movement of the wave 
timama [110]3 years ago
3 0

Explanation:

As ocean waves approach shore, their velocity decreases. The relation between the velocity, wavelength and the frequency of a wave is given by :

v=f\times \lambda

f = frequency

\lambda = wavelength

f=\dfrac{v}{\lambda}..............(1)

So, as the velocity of the wave increases, the wavelength and frequency also increases as per equation (1). When the velocity decreases, the frequency and the wavelength also decreases.

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8 0
3 years ago
Read 2 more answers
How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a po
miss Akunina [59]

Incomplete question as the charge density is missing so I assume charge density of 3.90×10^−12 C/m².The complete one is here.

An electron is released from rest at a distance of 0 m  from a large insulating sheet of charge that has uniform surface charge density 3.90×10^−12 C/m² .  How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?

Answer:

Work=1.06×10⁻²¹J

Explanation:

Given Data

Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²

Charge density σ=3.90×10⁻¹² C/m²

The electron moves a distance d=3.00×10⁻²m

Electron charge e=-1.6×10⁻¹⁹C

To find

Work done

Solution

The electric field due is sheet is given as

E=σ/2ε₀

E=\frac{3.90*10^{-12}C/m^{2}  }{2(8.85*10^{-12}C^{2} /N.m^{2} )}\\ E=0.22V/m

Now we need to find force on electron

F=eE\\F=(1.6*10^{-19}C )(0.22V/m)\\F=3.525*10^{-20}N

Now for Work done on the electron

W=F*d\\W=(3.525*10^{-20} N)(3.00*10^{-2}m)\\W=1.06*10^{-21}J

4 0
4 years ago
Ingrid is jogging at 9 km/h, she tosses a relay stick at 16 km/h to her teammate. How fast is the relay stick moving relative to
sesenic [268]

Answer:

27km/h

Explanation:

3 0
3 years ago
Light is shining in the images. Which object reflects a lot of light, absorbs a little light, and transmits almost no light?
SVEN [57.7K]
I believe it is B, the object that reflects a lot of light, absorbs a little light, and transmit almost no light.
6 0
3 years ago
Read 2 more answers
HELP: I’ve been stuck on this problem for a while now.
Arlecino [84]

Answer:

a.work done=force *displacement

=500N*46m

=23000 Joule

b.power=work done/time taken

=23000/25

=920 watt

c.GPE=m*g*h(m=mass,g=gravity due to acceleration,h=height)

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Explanation:

3 0
3 years ago
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