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xz_007 [3.2K]
4 years ago
13

A student look at ocean waves coming into the beach. An oceanwavd with more energy will

Physics
2 answers:
nlexa [21]4 years ago
8 0
It is A, they will have a greater height. I just had this question so it's right. 
pshichka [43]4 years ago
5 0
<span>C. Travel toward the beach faster</span>
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I need 1, 2 and 3 <br><br> Please help!
svetoff [14.1K]

1)Kenetic Energy is defined as energy which a body possesses by virtue of being in motion. 2)KE) is KE = 0.5 x mv2. Here m stands for mass, the measure of how much matter is in an object, and v stands for the velocity of the object, or the rate at which the object changes its position..

And I hope this helped :)

7 0
3 years ago
What is one latitude where there is no continental barriers?
AveGali [126]
The equator has no continental borders.
5 0
3 years ago
Read 2 more answers
Two 8.0 Ω lightbulbs are connected in a 12 V parallel circuit. What is the power of both glowing bulbs?
kati45 [8]

Answer:

96w

Explanation:

p=Iv..where v=12 and I=8.0

8 0
3 years ago
Please Help
musickatia [10]
The answer is C because <span>this movement is caused by the heat in Earth and creates </span>convection currents. <span>Convection currents in the asthenosphere cause movement of Earth's tectonic plates. </span> 
4 0
3 years ago
Read 2 more answers
At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise betw
allsm [11]

To develop this problem it is necessary to apply the concepts related to the Dopler effect.

The equation is defined by

f_i = f_0 \frac{c}{c+v}

Where

f_h= Approaching velocities

f_i= Receding velocities

c = Speed of sound

v = Emitter speed

And

f_h = f_0 \frac{c}{c+v}

Therefore using the values given we can find the velocity through,

\frac{f_h}{f_0}=\frac{c-v}{c+v}

v = c(\frac{f_h-f_i}{f_h+f_i})

Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1

v = 353(\frac{2.4-1}{2.4+1})

v = 145.35m/s

Therefore the cars goes to 145.3m/s

7 0
3 years ago
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