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Elenna [48]
3 years ago
13

What is the most significant challenge for organisms that live in estuaries?

Physics
1 answer:
marysya [2.9K]3 years ago
8 0

"Changing water salinity" is the most significant challenge for organisms that live in estuaries.

<u>Answer:</u> Option D

<u>Explanation:</u>

For estuaries, alkalinity levels are usually the maximum at a river's mouth where the ocean water falls for, and the minimum upstream where freshwater falls in. Although salinity vary throughout the tidal cycle. In estuaries, salinity rates usually decrease in spring as snow melt and rain raises the freshwater flow from streams and groundwater.

It influences the chemical environments within the estuary, especially the dissolved oxygen (DO) levels in the water. The level of oxygen that would get dissolved in water or its solubility get declined when the alkalinity rises.

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A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t
kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

W=Fd cos \theta (1)

where

F is the magnitude of the force

d is the distance covered by the object

\theta is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

\theta=0

And the distance covered is equal to the circumference of the circle, which is:

d=2\pi r=2\pi (2.5 m)=15.7 m

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

W=(3.0)(15.7)(cos 0)=47.1 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
If the mass of a material is 46 grams and the volume of the material is 14 cm^3, what would the density of the material be?
umka21 [38]

<u>Answer</u>

3.2857 g/cm³


<u>Explanation</u>

Density of a material is the mass of a unit volume of that material. It's SI unit is Kg/m³ but its has other units like g/cm³.

Density  = mass/volume

             = 46/14

             = 3.2857 g/cm³

3 0
3 years ago
Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the di
vekshin1

Here is the full question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

A = \pi (\frac{D}{2})^2

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

d= \frac{A}{N} =\frac{\pi (\frac{D}{2})^2 }{10, 000}

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

d = \pi r^2 e

r^2_e= \frac{d}{\pi}

r_e = \sqrt{\frac{d}{\pi} }

replacing d = \frac{\pi (\frac{D}{2})^2 }{10, 000} in the equation above; we have:

r_e = \sqrt{\frac{\frac{\pi (\frac{D}{2})^2 }{10, 000}}{\pi} }

r_e = \sqrt{\frac{(\frac{D}{2})^2 }{10, 000}}

r_e = \sqrt{\frac{(\frac{100,000}{2})^2 }{10, 000}}

r_e = 500 ly

The distance (s) between each civilization = 2(r_e)

= 2 (500 ly)

= 1000 light-years (ly)

4 0
3 years ago
A layer of oil (n = 1.38) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid.
Vinil7 [7]

Answer:

Refractive index of unknown liquid = 1.56

Explanation:

Using Snell's law as:

n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}

Where,  

{\theta_i}  is the angle of incidence  ( 65.0° )

{\theta_r} is the angle of refraction  ( 53.0° )

{n_r} is the refractive index of the refraction medium  (unknown liquid, n=?)

{n_i} is the refractive index of the incidence medium (oil, n=1.38)

Hence,  

1.38\times {sin65.0^0}={n_r}\times{sin53.0^0}

Solving for {n_r},

Refractive index of unknown liquid = 1.56

4 0
3 years ago
A ball is kicked from a height of 20 meters above the ground. If the initial velocity is 10 m/s, how long is the ball in flight
nasty-shy [4]
S=20 m
v=10 m/s
t=s/v
= 20/10
= 2 s.
8 0
3 years ago
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