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3 years ago

What are the subzones for each main zone?

Physics

1 answer:

Makovka662 [10]3 years ago

5 0

Answer:

The oceanic zone is subdivided into the epipelagic, mesopelagic, and bathypelagic zones on the basis of amount of light that reaches different depths. The mesopelagic (disphotic) zone, where only small amounts of light penetrate, lies below the Epipelagic zone.

Explanation:

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What do biologists, geologists, and physicists have in common, how are they differnet

Brilliant_brown [7]

They all study under some sect of science.
And they all study under different sects of science ie. Biologist study of life Geologists study of Geology, and Physicists study of physics.

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4 years ago

Read 2 more answers

Why does the gravitational field strength increase with latitude

otez555 [7]

The second major reason for the difference in gravity at differentlatitudes is that the Earth's equatorial bulge (itself also caused by centrifugalforce from rotation) causes objects at the Equator to be farther from the planet's centre than objects at the poles.

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3 years ago

una onda longitudinal tiene una frecuencia de 200 hz y una longitud de onda de 4.2m ¿cual es la rapidez de la onda?

swat32

Answer:

v = 8.4 m/s

Explanation:

The question ays, "A longitudinal wave has a frequency of 200 Hz and a wavelength of 4.2m. What is the speed of the wave?".

Frequency of a wave, f = 200 Hz

Wavelength = 4.2 cm = 0.042 m

We need to find the speed of the wave. The formula for the speed of a wave is given by :

$v=f\lambda\\\\v=200\times 0.042\\\\=8.4\ m/s$

So, the speed of the wave is equal to 8.4 m/s.

4 0

3 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at

inn [45]

Answer:

a) $v=13.2171\,m.s^{-1}$

b) $H=8.9605\,m$

Explanation:

Given:

mass of bullet, $m=4.97\times 10^{-3}\,kg$

compression of the spring, $\Delta x=0.0476\,m$

force required for the given compression, $F=9.12 \,N$

(a)

We know

$F=m.a$

where:

a= acceleration

$9.12=4.97\times 10^{-3}\times a$

$a\approx 1835\,m.s^{-2}\\$

we have:

initial velocity, $u=0\,m.s^{-1}$

Using the eq. of motion:

$v^2=u^2+2a.\Delta x$

where:

v= final velocity after the separation of spring with the bullet.

$v^2= 0^2+2\times 1835\times 0.0476$

$v=13.2171\,m.s^{-1}$

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

$u=13.2171\,m.s^{-1}$

∵At maximum height the final velocity will be zero

$v=0\,m.s^{-1}$

Using the equation of motion:

$v^2=u^2-2g.h$

where:

h= height

g= acceleration due to gravity

$0^2=13.2171^2-2\times 9.8\times h$

$h=8.9129\,m$

is the height from the release position of the spring.

So, the height from the latched position be:

$H=h+\Delta x$

$H=8.9129+0.0476$

$H=8.9605\,m$

4 0

4 years ago

a baseball is pitched with a speed of 35.0m/s. how long does it take the ball to travel 18.4 m from the pitchers mound to home p

Vlad [161]

Given the speed and the distance, to find time you can use the formula speed is equal to distance over time. From there you can manipulate the equation for time to equal the distance divided by speed. Time is equal to 18.4 meters divided by 35m/s which equals 0.526 seconds.

7 0

3 years ago