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Leto [7]
3 years ago
11

What kind of water temperature lowers the dissolved oxygen

Physics
2 answers:
bija089 [108]3 years ago
6 0
<h2>Answer:</h2>

<u>Warm water temperature </u><u>lowers the dissolved oxygen</u>

<h2>Explanation:</h2>

Water temperature is one of the most important characteristics of an aquatic system. It affects the solubility of oxygen to a great extent. The solubility of oxygen decreases as water temperature increases.The reason for this inverse relationship between dissolved oxygen and temperature is that the solubility of a gas in a liquid is an equilibrium phenomenon therefore cold water can hold more dissolved oxygen than warm water.

ehidna [41]3 years ago
5 0
<h3>Answer</h3>

At a high temperature above 20° oxygen solubility starts to decrease.

<h3>Explanation</h3>

Oxygen, O2 is a very essential component of water as we can see in its chemical formula h2O.

The solubility of oxygen decreases as temperature increases. This means that warmer water will have less dissolved oxygen than does cooler water.

<h3>Other factors that affects oxygen solubility in water</h3>

Salt levels

higher the salt levels in water, lower will be oxygen in it.

Pressure

Water at lower altitudes can hold more dissolved oxygen than water at higher altitudes because dissolved oxygen will increase as pressure increases.

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In short, Your Answer would be "True"

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5 0
3 years ago
B)
Vlad1618 [11]

Answer:

a) 16m/s b) 192m

Explanation:

v1=32m/s a=-2m/s^2 t=8s v2=? d=??

a) I will use this equation v2= v1 + a*t

v2= 32m/s + -2m/s^2 * 8s

v2= 32m/s + -16m/s

v2= 16m/s

b) v2^2=v1^2 + 2ad

rearranging

v2^2-v1^2=2ad

v2^2-v1^2/2= a d

v2^2-v1^2/2a=d

16m/s^2 - 32m/s^2/ 2 x-2m/s^2 =d

d=192m

5 0
2 years ago
A car going 22 m/s increases its speed to pass a truck. Five seconds later the car is going 35 m/s. Calculate the acceleration o
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I don't know

Explanation:

4 0
3 years ago
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The suspension system mounts the car's wheels<br> solid on the frame. True or false?
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3 years ago
On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the
IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = \frac{1}{2}\varepsilon _{0}E^{2}

          =\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119

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(b) energy density of magnetic field = \frac{B^{2}}{2\mu _{0}}

=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0
3 years ago
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