<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.
<h2>
Power is 11 W</h2>
Explanation:
Power = Work ÷ Time
Work = Force x Displacement
Force = 22 N
Displacement = 3 m
Time = 6 seconds
Substituting
Work = Force x Displacement
Work = 22 x 3 = 66 J
Power = Work ÷ Time
Power = 66 ÷ 6
Power = 11 W
Power is 11 W
Answer:
Explanation:
The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .
Their potential energy at distance d
= kq₁q₂ / d
= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J
= 16.2 J
Their total kinetic energy will be equal to this potential energy.
2 x 1/2 x mv² = 16.2
= 3 x 10⁻⁶ v² = 16.2
v = 5.4 x 10⁶
v = 2.32 x 10³ m/s
When masses are different , total P.E, will be divided between them as follows
K E of 3 μ = (16.2 / 30+3) x 30
= 14.73 J
1/2 X 3 X 10⁻⁶ v₁² = 14.73
v₁ = 3.13 x 10³
K E of 30 μ = (16.2 / 30+3) x 3
= 1.47 J
1/2 x 30 x 10⁻⁶ x v₂² = 1.47
v₂ = .313 x 10³ m/s
The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N
<h3>Data obtained from the question </h3>
- Initial velocity (u) = 8.5 m/s
- Final velocity (v) = 7.5 m/s
- Time (t) = 5 ms = 0.25 s
- Mass (m) = 2.5 Kg
- Force (F) = ?
<h3>How to determine the force</h3>
The force exerted on the ball can be obtained as follow:
F = m(v + u) / t
F = [2.5(7.5 + 8.5)]/ 0.25
F = 40 / 0.25
F = 160 N
Thus, the force exerted on the ball is 160 N
Learn more about momentum:
brainly.com/question/250648
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Answer:
0.5 Hz
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