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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.

You might be interested in

what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three

grin007 [14]

Answer:

Approximately $5.11 \times 10^{-19}\; {\rm J}$ .

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: $R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}$ .

Look up the speed of light in vacuum: $c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}$ .

Look up Planck's constant: $h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}$ .

Apply the Rydberg formula to find the wavelength $\lambda$ (in vacuum) of the photon in question:

$\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}$ .

The frequency of that photon would be:

$\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}$ .

Combine this expression with the Rydberg formula to find the frequency of this photon:

$\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}$ .

Apply the Einstein-Planck equation to find the energy of this photon:

$\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}$ .

(Rounded to three significant figures.)

6 0

2 years ago

Can somebody definee newton1 word to word pls

Dimas [21]

Answer:

One newton is a unit of force equal to the force needed to move a one kilogram mass by one meter per second per second. Word originsThis word is named for Sir Isaac Newton (1642-1727), the great English mathematician and physicist who discovered gravity.

Explanation:

hope this help you

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7 0

3 years ago

How much work done when .0080 C is moved through a potential difference of 1.5 V? Use W = qV. A.

grin007 [14]

Answer:

0.012 J

Explanation:

We are given:

q = 0.0080C

Potential difference = 1.5V

W=qV

Substituting the values into the equation:

W=0.0080*1.5= 0.012J

8 0

3 years ago

A 2200-kg railway freight car coasts at 4.1 m/s underneath a grain terminal, which dumps grain directly down into the freight ca

777dan777 [17]

Answer:

The answer is " $2.41 \times 10^3$ "

Explanation:

Given:

$m_i = 2000 \ kg \\\\v_i= 4.1 \ \frac{m}{s} \\\\v_f = 3.4 \ \frac{m}{s} \\$

Using formula:

$\to m_iv_i = m_fv_f \\\\\to m_f= \frac{m_iv_i}{v_f}$

$p_i, p_f$ = system initial and final linear momentum.

$V_i, v_f$ = system original and final linear pace.

$m_i$ = original weight of the car freight.

$m_f$ = car's maximum weight

$= \frac{ 2000 \times 4.1}{3.4}\\\\= \frac{ 8.2\times 10^3}{3.4}\\\\= 2.41 \times 10^3$

$\boxed{m_f = 2.41 \times 10^3}$

8 0

2 years ago

If, as is typical, each of them breathes about 500 cm3 of air with each breath, approximately what volume of air (in cubic meter

deff fn [24]

Answer:

a) 12614.4 m^3

b) 28.8 m

Explanation:

The complete question is

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm^3 of air with each breath, approximately what volume of air (in cubic meters) do these astronauts breathe in a year? (b) What would the diameter (in meters) of the space station have to be to contain all this air?

The average breathing rate is 12 breaths per minute

there are 60 minutes x 24 hours x 365 days in a year = 525600 minutes in a year

if an average human takes 12 breath per minute, then in a year an average human take 12 x 525600 = 6307200 breath

For the four astronauts, the amount of breath = 4 x 6307200 = 25228800 breath in total.

The volume of air per breath = 500 cm^3

1 cm^3 = 10^-6 m^3

500 cm^3 = $x$ m^3

$x$ = 500 x 10^-6 = 5 x 10^-4 m^3

Therefore in a year, the volume of these astronauts breath in a year = 5 x 10^-4 x 25228800 = 12614.4 m^3

b) If the space station is spherical, the volume will be given as = $\frac{4}{3} \pi r^3$

where r is the radius of the space station

This volume of the space station must be able to contain all the volume of breath produced by the astronauts which is = 12614.4 m^3

Equating, we have

12614.4 = $\frac{4}{3} \pi r^3$

12614.4 = $\frac{4}{3}*3.142*r^3$

12614.4 = 4.1893 $r^{3}$

$r^{3}$ = 12614.4/4.1893 = 3011.1

r = $\sqrt[3]{3011.1}$ = 14.4 m

diameter of the space station = 14.4 m x 2 = 28.8 m

3 0

3 years ago