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ivolga24 [154]
2 years ago
5

What happens to waves near the shore?

Physics
1 answer:
Lady_Fox [76]2 years ago
8 0

Answer As a wave travels across the open ocean, it gains speed. When a wave reaches a shallow coastline, the wave begins to slow down due to the friction caused by the approaching shallow bottom. ... Think of it like driving a car at high speed and then slamming on the breaks. Everything is going to fly to the front.:Waves at the Shoreline: As a wave approaches the shore it slows down from drag on the bottom when water depth is less than half the wavelength (L/2). The waves get closer together and taller. ... Eventually the bottom of the wave slows drastically and the wave topples over as a breaker. hope this helps have a nice night❤️❤️❤️

Explanation:

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who brought astronomy out of the dark ages by declaring: "finally we shall place the sun himself at the center of the universe.
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Nicolaus Copernicus was the one who brought astronomy out of the dark ages.

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To know more about the Heliocentric system refer to the link brainly.com/question/3491738?referrer=searchResults.

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2 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

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ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

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(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

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Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

b. Time Taken to complete 30revolution

θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

Then, t=120π/58π

t=2.07seconds

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wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0
3 years ago
Read 2 more answers
The superheroine Xanaxa, who has a mass of 68.1 kg , is pursuing the 75.3 kg archvillain Lexlax. She leaps from the ground to th
Alexandra [31]

The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

U = mg\Delta h

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g = Acceleration due to gravity

The change in height will be,

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h_f = -16.1m

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\Delta h = -16.1m-0m

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4 0
3 years ago
A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh
Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

F_s = F_o [\frac{v}{v_s + v} ] \\\\

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0
3 years ago
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