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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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A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an
posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle \theta=30°

We need to calculate the angle in radian

\theta=30\times\dfrac{\pi}{180}

\theta=0.523\ rad

We need to calculate the path length

Using formula of path length

Path\ length =angle\times radius

Path\ length=0.523\times5.9

Path\ length =3.09\ m

(b). Angle = 30 rad

We need to calculate the path length

Path\ length=30\times5.9

Path\ length=177\ m

(c). Angle = 30 rev

We need to calculate the angle in rad

\theta=30\times2\pi

\theta=188.4\ rad

We need to calculate the path length

Path\ length=188.4\times5.9

Path\ length =1111.56\ m

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

8 0
3 years ago
momentum A proton interacts electrically with a neutral HCl molecule located at the origin. At a certain time t, the proton’s po
arlik [135]

Answer:

\ m/s

Explanation:

F = Force =

m = Mass of proton = 1.7\times 10^{-27\ kg

t = Time taken = 2\times 10^{-14}\ s

Acceleration is given by

a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{}{1.7\times 10^{-27}}\\\Rightarrow a=\ m/s^2

v=u+at\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\\\Rightarrow v=\ m/s

The velocity of the proton is \ m/s

6 0
3 years ago
a racing car undergoes a uniform acceleration of 4.00m/s2. if the net force causing the acceleration is 3.00 times 10^3 N, what
yKpoI14uk [10]

Answer: 750Kg

Explanation:

Recall that force is the product of the mass M, of an object moving at a uniform acceleration.

i.e Force = Mass x Acceleration

In this case, Mass = ?

Force = 3.00 x 10^3 N = (3.00 x 1000N)

= 3000N

Uniform acceleration = 4.00m/s^2

Force = Mass x Acceleration

3000N = Mass x 4.00m/s^2

Mass = (3000N/4.00m/s^2)

Mass = 750Kg (The SI unit of mass is kilograms)

Thus, the mass of the car is 750Kg

4 0
3 years ago
List and Explain three ways study groups benefit your learning
vodomira [7]

Answer:

The answer

Explanation:

Thinking together, Better friendship, Makes teacher happy.

4 0
3 years ago
Read 2 more answers
Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

8 0
3 years ago
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