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3 years ago

The increase in height as a tsunami approaches shore is due to

Physics

2 answers:

ryzh [129]3 years ago

5 0

Answer:

C. The decrease in speed as the wave approaches shore.

Explanation:

The waves break when approaching the shore because the depth decreases. Thus, the wave travels more slowly and increases its height. There comes a time when the part of the wave on the surface travels faster than the one that travels under water, the ridge destabilizes and falls against the ground.

OleMash [197]3 years ago

3 0

Answer:

Explanation:

A tsunami is a series of successive ocean waves usually caused by landslide, earthquake or volcanic eruption. The speed of a tsunami decreases and its height increases as it moves from the deep part of a water body to the shallow part (e.g shore) of the water body. In other words, for a tsunami, the deeper the water, the slower its speed and consequently the higher it grows. Therefore it is safe to say that an increase in the height of a tsunami as it approaches a shore is due to the decrease in speed as the wave approaches the shore.

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noname [10]

Answer:

0.3125m/s

Explanation:

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3 years ago

A player kicks a football from ground level with an initial velocity of 27.0 30.0 above the horizontal Find the ball's maximum h

stepladder [879]

Answer:

Given

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Force of gravidity g) =9.8

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3 years ago

Supposing d(t) is known to have value D,

creativ13 [48]

Answer:

The procedure is: solve the quadratic equation for $t$ .

Explanation:

This question assumes uniformly accelerated motion, for which the distance d a particle travels in time t is given by the general equation:

$d(t)=d_0+v_0t+at^2/2$

That is a quadratic equation, where the independent variable is the time $t$ .

Thus, the procedure that will find the time t at which the distance value is known to be D is to solve the quadratic equation for $t$ .

To solve it you start by changing the equation to the general form of the quadratic equations, rearranging the terms:

$(a/2)t^2+v_0t+(d_0-D)=0$

Some times that equation may be solved by factoring, and always it can be solved by using the quadratic formula:

$t=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

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3 years ago

Proposed Exercise - Circular Movement

notka56 [123]

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

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3 years ago

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viktelen [127]

This question has already previously been answered. :)

Here it is: brainly.com/question/2141424

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3 years ago