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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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4. A student measures a temperature several times. The readings lie between 29.6 and 30.2 K. This
Alona [7]

Answer:

(C) apparently written incorrectly - it should be 29.9 +- .3 K

(read 29.9 plus or minus .3 K)

8 0
3 years ago
2.
krek1111 [17]

Answer:

c

Explanation:

a vector quantity has both magnitude and direction

6 0
3 years ago
PLEASE HELP!
sergiy2304 [10]

Answer: b

Explanation:

p=mv

p=30

m=1500

rearrange to find v

(30)=(1500)(v)

v= 30/1500

v = 0.02 m/s

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3 years ago
Which of the following statements apply to electric charges? please help this is the only question i need done!!
disa [49]

The only correct statement on that list is <em>choice-C</em>:  If a positively charged rod is brought close to a positively charged object, the two objects will repel.

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3 years ago
Read 2 more answers
A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an init
AURORKA [14]

a) 19.4 cm

b) 3.2 m/s

Explanation:

a)

A horizontal spring-mass system has a motion called simple harmonic motion, in which the mass oscillates following a periodic function (sine or cosine) around an equilibrium position.

As the system oscillates back and forth, its total mechanical energy (sum of elastic potential energy and kinetic energy) will remain conserved (since we consider friction negligible). The elastic potential energy at any point is given by:

U=\frac{1}{2}kx^2

where

k is the spring constant

x is the displacement of the system

While the kinetic energy at any point is

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

So the total mechanical energy of the system is

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

For this system, when it is initially released,

m = 0.6 kg

k = 165 N/m

x = 7 cm = 0.07 m

v = 3 m/s

So the total energy is

E=\frac{1}{2}(0.6)(3)^2+\frac{1}{2}(165)(0.07)^2=3.1 J

Since friction is negligible, this total energy remains constant. Therefore, when the system reaches its maximum stretch during the motion, the kinetic energy will be zero and all the mechanical energy will be elastic potential energy; so we will have:

E=U=\frac{1}{2}kx_{max}^2

where x_{max} is the maximum stretch. Solving for x_{max},

x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(3.1)}{165}}=0.194 m

So, 19.4 cm.

b)

The maximum speed in a spring-mass oscillating system is reached when the kinetic energy is maximum, and therefore, since the total energy is conserved, when the elastic potential energy is zero:

U=0

which means when the displacement is zero:

x = 0

So, when the system is transiting through the equilibrium position.

Therefore, the total mechanical energy is equal to the maximum kinetic energy:

E=K=\frac{1}{2}mv_{max}^2

where

m is the mass

v_{max} is the maximum speed

Here we have:

E = 3.1 J

m = 0.6 kg

Therefore, solving for the maximum speed,

v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(3.1)}{0.6}}=3.2 m/s

6 0
3 years ago
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