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larisa86 [58]
3 years ago
5

Minute after minute, hour after hour, day after day, ocean waves continue to splash onto the shore. Explain why the beach is not

completely submerged and why the middle of the ocean has not yet been depleted of its water supply. A) Ocean waves can only bring water to the shore but cannot take it back. The momentum of the water particles switches direction when the wave hits the shore and this momentum takes all the water back into the ocean. B) Ocean waves bring the water and also take it all the way back into the ocean. As such, water does not pile up on the beach. C) Ocean waves can only bring energy to the shore; the particles of the medium (water) simply oscillate about their fixed position. D) None of the choices are correct.
Physics
1 answer:
MAVERICK [17]3 years ago
4 0

Answer: C. Ocean waves can only bring energy to the shore; the particles of the medium (water) simply oscillate about their fixed position.

Explanation:

The reason why the beach is not completely submerged and the reason why the middle of the ocean has not been depleted of its water supply is due to the fact that water isn't transported by ocean waves.

It should be noted that even a single drop of water cannot be brought by the ocean wave to the shore from the middle of the ocean.

The only thing that the ocean waves can bring to the shore is energy. Hemce, the water particles oscillate in their fixed position which is vital in making sure that the beach isn't piled up with water.

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An inductor in an LC circuit has a maximum current of 2.4 A and a maximum energy of 56 mJ.
Harrizon [31]

Answer:

The energy stored in the capacitor, when the current in the inductor is 1.2 A, is 41.6 mJ.

Explanation:

In a LC oscillating circuit, the energy is stored in the electric field (between the plates of the capacitor) and in the magnetic field (surrounding the wires of the inductor).

At any time, the sum of both energies can be expressed as follows:

E = 1/2 Q² / C   +  1/2 L I²

In this type of circuit, energy oscillates, which means that it is exchanging between both fields all time.

When the capacitor is completely discharged, all the energy is stored in the magnetic field, and at that time, the current is maximum.

The total energy, when I is maximum, can be written as follows:

E = 1/2 L I² (1)

In our case, when I= 2.4A, E= 56 mJ.

So, we can find out the value of L, which will allow us to know the value of the magnetic energy at any time, having the value of the instantaneous current.

Solving for L in (1):

L = 2 *.56 mJ / (2.4)² A² = 20 mH

The next step is getting the value of the energy stored in the inductor, when I = 1.2 A, as follows:

Em = 1/2 *20 mH.* (1.2)² A² = 14.4 mJ

As the total energy must be always the same, i.e., 56 mJ, the energy stored in the capacitor, assuming no losses, must be the difference between the total energy and the one stored in the magnetic field:

Ec = 56 mJ - 14.4 mJ = 41.6 mJ

3 0
3 years ago
Which best describes the energy of a sound wave as it travels through a medium?
qaws [65]

Answer:

it increases

Explanation:

7 0
3 years ago
Read 2 more answers
In a 49 s interval, 595 hailstones strike a glass window of an area of 0.954 m at an angle of 25° to the window surface. Each ha
eduard

Average  force on the window: 0.32 N

Explanation:

The average force exerted on the window is given by Newton's second law

F=\frac{\Delta p}{\Delta t}

where

\Delta p is the net change in momentum of the hailstones in a time interval of \Delta t

In order to find the change in momentum, we have to consider only the component of the hailstone's momentum perpendicular to the window, therefore:

p_i =m u sin \theta is the initial momentum of one hailstone, with

m = 7 g = 0.007 kg is the mass

u=4.5 m/s is the initial speed

\theta=25^{\circ} is the angle with the window

The final momentum is

p_f = mv sin \theta

where

v = 4.5 m/s is the final speed (the  collision is elastic so the speed is equal, while the direction changes)

\theta=-25^{\circ} (after the rebound, the direction has changed)

So the change in momentum of 1 hailstone is

\Delta p = mv sin(-25^{\circ})-mu sin(25^{\circ})=-2mu sin(25^{\circ})=-0.0266 kg m/s

We are interested only in the magnitude, so

\Delta p = 0.0266 kg m/s

There are 595 hailstones hitting the window in 49 s, so the total change in momentum is

\Delta p = 595\cdot 0.0266 = 15.8 kg m/s

And therefore, the average force on the window is

F=\frac{\Delta p}{\Delta t}=\frac{15.8}{49}=0.32 N

Learn more about  force:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

3 0
3 years ago
a distant galaxy is studied with a radio telescope x ray telescope and optical light telescope. the images form which set of mus
aivan3 [116]

Answer:

The modern instruments or we can say the different levels of telescopes are used to explore and study the distant galaxies. i.e the Hubble telescope is out there providing the data regarding the different properties of the celestial entities which in other case is not visible to the human naked eye.

Explanation:

  • Scientists and research workers are in constant search for more answers as they explore the universe and implement the laws of physics on the celestial entities. But, most of the objects inside the universe are not visible to human naked eye, as they are far from sight and thus more advanced form of instruments like the x-ray, optical, and light telescopes are used to determine the different properties of the celestial entities inside the universe.
  • As, these telescopes includes the most recent "Hubble telescope", which is out there inside the space to explore the universe and more over the galaxies by subjecting them with x-rays and then provide us with a very rough but valid results to study the distant galaxies.

6 0
3 years ago
A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.
Anastasy [175]

Initially, the velocity vector is \langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 4.9(0.2)^2, so the velocity is \langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle.

Converting back to direction and magnitude, we get \langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle

4 0
3 years ago
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