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3 years ago

When do surface waves reach land

Physics

2 answers:

Alex_Xolod [135]3 years ago

7 0

Answer:

Surface waves are the kind of waves that occur on the surface of the ocean. The reason behind their occurrence is the geological effects, usually in the form of winds in near geographical regions or in that region.

These waves reach the land when the geological winds give a push to the waves and they travel miles and finally hit the land. These waves range in sizes, for small sizes to the the size of tsunami.

fiasKO [112]3 years ago

3 0

Answer with Explanation:

"Surface waves" are also known as "wind-generated waves." Such waves occur on the surface of the water that has a zero parallel shear stress. Their speed upon reaching the land will depend upon the wind blowing over the fluid's surface. So, this means that they could travel several miles before actually reaching the land. They could also come in different sizes, from small ripples to big tsunamis.

The factors that tell the time when the surface waves reach the land are: wind speed, depth of the water, wind duration, the area that is affected by the fetch (width) and the uninterrupted distance of open water.

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g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

3 years ago

An eccentric inventor attempts to levitate a cork ball by wrapping it with foil and placing a large negative charge on the ball

djverab [1.8K]

Answer:

because of the idea that like charges get repulsion as a force.

Explanation:

because you wrap the ball with foil, the negative charges will leave the foil and go into the ball by induction. This leaves the foil as a positively charged particle since its electrons left it for the ball, making the ball a negatively charged particle. but if you bring the negative charge near the foil, the electrons will transfer from that and go into the foil, making it negatively charged. Now, because the ball and the foil have the same charge, they repel. the foil flies off.

7 0

2 years ago

(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o

IgorLugansk [536]

Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

$W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J$