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lidiya [134]
3 years ago
12

Why is the deep ocean cold and dark?

Physics
2 answers:
Tomtit [17]3 years ago
8 0
Because no sunlight can penetrate it
Valentin [98]3 years ago
3 0

WHY THEY ARE DEEP>>>>

according to: amnh.com

the oceans are deep because it states that  "The ocean is very, very deep; light can only penetrate so far below the surface of the ocean. As the light energy travels through the water, the molecules in the water scatter and absorb it. At great depths, light is so scattered that there is nothing left to detect."

WHY THEY ARE COLD>>>

according to: ocean service.com

oceans are cold because it states that  "Cold water has a higher density than warm water. Water gets colder with depth because cold, salty ocean water sinks to the bottom of the ocean basins below the less dense warmer water near the surface"

WHY THEY ARE DARK>>>

according to amnh.com

Oceans are dark because it states that "The ocean is very, very deep; light can only penetrate so far below the surface of the ocean. As the light energy travels through the water, the molecules in the water scatter and absorb it."

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I just need x isolated
mamaluj [8]

Answer:

x=\frac{-y+\sqrt{y^2+4rt} }{2r}

x=\frac{-y-\sqrt{y^2+4rt} }{2r}

Explanation:

rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0

Solve using the quadratic formula.

x=\frac{-y+\sqrt{y^2+4rt} }{2r}

x=\frac{-y-\sqrt{y^2+4rt} }{2r}

7 0
3 years ago
Read 2 more answers
What would we see looking across a flat land of the earth was curved?
Scrat [10]
You would see mountains off in the distance as if the earth was actually flat. but it seems flat because its so big
7 0
3 years ago
A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collis
Elza [17]

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2               ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1                  ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

4 0
3 years ago
What is the relationship between temperature and height in the troposphere ,stratosphere,mesophere,thermosphere,exosphere
Soloha48 [4]

In general the higher we go the cooler it becomes . temperature fall continues within the lowermost layer of our atmosphere, known as the troposphere. Above the troposphere, the stratosphere exists and in this region the temperature increases with an increase in altitude. The mesosphere is the part of the atmosphere that lies above the stratosphere. In this region, the temperature falls with an increase in altitude. And finally as we continue rising up, we reach the thermosphere, where the temperature increases with increased altitude.

3 0
3 years ago
A stone is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 20 m/s. Th
Hitman42 [59]

Answer:

Explanation:

Given

inclination \theta =30^{\circ}

initial speed u=20\ m/s

Point of release is 45 m above the ground

Considering stone to be a projectile, so time taken by projectile for its zero vertical displacement is

t_1=\frac{2u\sin \theta }{g}

t_1=\frac{2\times 20\times \sin 30}{10}

t_1=2\ s

Now after completing zero vertical displacement , stone needs to travel another 45 m in downward direction with initial speed u=20\sin 30

h=u_yt+\frac{1}{2}a_yt^2

where, h=height

u_y=vertical velocity

a_y=vertical acceleration

t_0=time

45=20\sin 30+\frac{1}{2}(9.8)(t_0)^2

t_0^2=\frac{70}{9.8}

t_0=2.64\ s

thus total time time required is t=t_0+t_1=2.64+2=4.64\ s

vertical velocity just before hitting

v_y=\sqrt{u_y^2+2\times a_y\times s}

v_y=\sqrt{10^2+2\times 10\times 45}

v_y=\sqrt{1000}=31.622\ m/s

Horizontal velocity v_x=u\cos 30=17.32\ m/s

Net velocity Just before hitting =\sqrt{v_x^2+v_y^2}

=\sqrt{(17.32)^2+(31.62)^2}

=\sqrt{1299.82}=36.05\ m/s

                 

7 0
3 years ago
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