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3 years ago

Earthquakes that originate beneath the ocean floor produce huge tidal waves called _____.

Physics

2 answers:

Vinvika [58]3 years ago

6 0

Answer:

tsunamis

Explanation:

ololo11 [35]3 years ago

3 0

Answer;

-Tsunami

Explanation;

-Tsunami is a series of large ocean waves (or "wave train") of extremely long wavelength and period, usually generated when a gigantic body of water, such as an ocean, is suddenly displaced on a massive scale by an underwater disturbance such as an earthquake occurring on or near the sea floor or a volcanic eruption.

-After a sudden displacement of a large water volume by seismic activity (earthquake), the ocean floor is raised or dropped and large tsunami waves can be formed by gravitational forces.

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The human eye can readily detect wavelengths from about 400 nm to 700 nm. If white light illuminates a diffraction grating havin

nadya68 [22]

Answer:

The range of angles is from 17.50° to 31.76°

Explanation:

The diffraction grid equation is as follows:

$dsen\theta=m\lambda$

Clearing for $\theta$

$sen\theta=\frac{m\lambda}{d}$

$\theta=sen^{-1}(\frac{m\lambda}{d})$

where $\theta$ is the angle, $m$ is the order, in this case $m=1$ , $\lambda$ is the wavelength, and $d$ is defined as follows:

$d=\frac{1}{resolution}$

and since the resolution is 750 lines/mm wich is the same as $750lines/1x10^{-3}m$

$d$ will be:

$d=\frac{1}{750lines/1x10^{-3}m}=\frac{1x10^{-3}m}{750lines}=1.33x10^{-6}m$

wich is the distance between each line of the diffraction grating.

substituting the values for $m$ and $d$ :

$\theta=sen^{-1}(\frac{(1)\lambda}{(1.33x10^{-6}m)})$

And we need to find two angle values: one for when the wavelength is 400nm and one for when it is 700 nm. So we will get the angle range

$\theta=sen^{-1}(\frac{(400x10^{-9})}{(1.33x10^{-6}m)})=17.50$

and

$\theta=sen^{-1}(\frac{(700x10^{-9})}{(1.33x10^{-6}m)})=31.76$

The range of angles is from 17.50° to 31.76°

3 0

3 years ago

Which parameter of a projectile depends on the horizontal as well as the vertical component of velocity of projection?

jekas [21]

Just about anything you could ask about a projectile AFTER it's launched
depends on both components of the launch velocity.

Here are some that I can think of:

-- angle of launch
-- magnitude of launch velocity
-- location at any time after launch
-- magnitude of velocity at any time after launch
-- direction of velocity at any time after launch
-- distance of the landing point from the launch point

6 0

3 years ago

Read 2 more answers

Tp-18 what should you do when operating a boat in large waves and high wind?

deff fn [24]

1) Try to head into the waves at some slight angle and the speed of the boat should be reduced.

2) In order to ride up and over the waves, the speed of the boat should be slow.

3) The less the speed of the boat, and the less strain will be put on the hull and superstructure.

5 0

3 years ago

Ultraviolet rays from the sun are able to reach Earth's surface because A. They require air to travel through B. They have less

9966 [12]

Yeah, because of it's short frequencies, ultraviolet rays can travel through empty space- D
: D

3 0

3 years ago

11. A disk 8.00 cm in radiu: rotates at a constant rate of 1200 revinin about its central axis Determine (a) its angular speed i

marshall27 [118]

Answer:

A disk 8.00 cm in radiu: rotates at a constant rate of 1200 revinin about its central axis Determine (a) its angular speed in zadians per second, (b) the tangential speed at a point 3.00 cm from its center, (c) the radial aceleration of a point on the rim, and (d) the total distance a point our tke rim noves ign-2.00 s (E) The moment of inertia if it's mass is 2Kg? is the answer

5 0

3 years ago