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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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The fulcrum of a first-class lever divides its 9.0 m arm into two sections—a 6.0 m arm and a 3.0 m arm. You place a rock weighin
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For balancing the lever, force on both the sides shall be equal. so,
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2 years ago
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
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Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

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    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

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