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3 years ago

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ANSWER ASAP Which factor determines the amount of gas that can dissolve in ocean water? (A) the motion of the air above the ocea

n water (B) the temperature of the air above the ocean water (C) the motion of the ocean water currents D the temperature of the ocean water

2 answers:

Andru [333]3 years ago

6 0

Answer:

The amount of each gas that can dissolve in the ocean depends on the solubility and saturation of the gas in water. Solubility refers to the amount of a dissolved gas that the water can hold under a particular set of conditions, which are usually defined as 0o C and 1 atmosphere of pressure.

Explanation:

hope this helps

Westkost [7]3 years ago

5 0

Answer: the motion of the air above the ocean water

Explanation:

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3 years ago

An L-R-C series circuit is connected to a 120 Hz ac source that has Vrms = 82.0 V. The circuit has a resistance of 71.0 Ω and an

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Explanation:

Given that,

Frequency of LCR circuit is 120 Hz

RMS voltage, $V_{rms}=82\ V$

Resistance of circuit, R = 71 Ω

Impedance, Z = 107 Ω

We need to find the average power is delivered to the circuit by the source. Firstly, finding the rms value of current,

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4 years ago

Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

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Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago