What is the cause of most waves and how can this change waves' characteristics?

Physics

2 answers:

Tanzania [10]3 years ago

8 0

There are three main factors that affect wave formation: wind velocity, fetch, and duration.

Waves are most commonly caused by wind. Wind-driven waves, or surface waves, are created by the friction between wind and surface water. As wind blows across the surface of the ocean or a lake, the continual disturbance creates a wave crest.

AveGali [126]3 years ago

6 0

To answer this you must first know the definition of a wave. A wave is an oscollation (something that goes back and forth repeatedly) in a medium. A medium is something that the wave travels through, for example, the sound that we hear travels through air, that's to say air is the medium. The cause of a wave is usually caused by introducing energy into a medium. for example, if you shake an electron you introduce energy into the electromagnetic medium. Or if you smack your hands together you introduce energy into the air medium. The energy you introduced travels through the medium. After you shook the electron you gave the electromagnetic medium energy.

Hope it helped

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on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker

DIA [1.3K]

Momentum
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.

Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum

Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum

Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude

!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v [solve for v]
v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s

If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]

The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.

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3 years ago

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Klio2033 [76]

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3 years ago

At a local swimming pool, the diving board is elevated 21.5 ft above the pool's surface and overhangs the pool edge by L = 6 ft.

Galina-37 [17]

Answer:

Part (a) $t_w=\sqrt{\dfrac{2h}{g}}$

Part (b) 1.16 s

Part (c) 21.3 ft

Explanation:

Part (a) There are two motions involved: a horizontal motion and a vertical motion. The vertical motion is under gravity and is irrespective of the speed of the diver because the speed is horizontal. Also, the horizontal motion is not accelerated. Therefore, both motions take the same time $t_w$ which is determined by the vertical motion.

With respect to the vertical motion, there is no initial speed because the speed was fully horizontal. In fact, the diver could as well have dropped at the edge of the board and he would spend the same time to get to the pool surface.

Initial speed, u = 0 ft/s

Acceleration, a = g (it is purely under gravity)

Distance, s, = h

Time, $t_w$

Applying one of the equations of motion which has the required parameters,

$s=ut_w +\frac{1}{2}at_w^2$

$h=0 +\frac{1}{2}gt_w^2$

$t_w^2=\frac{2h}{g}$

$t_w=\sqrt{\dfrac{2h}{g}}$

Part (b) Substituting the values, taking g = 32.2 ft/s ${}^2$

$t_w=\sqrt{\dfrac{2\times24. 5}{32.2}}=\sqrt{1.335}=1.16$ s

Part (c) Total horizontal distance = horizontal distance due to diving + overhang (L = 6)

Horizontal distance due to diving is dependent solely on the horizontal component of the motion. This component has no acceleration, that is, has uniform speed of 13.2 ft/s.

This distance = speed $\times$ time