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allochka39001 [22]
3 years ago
7

A high-pressure system is moving into an area. Which type of weather can people in this area expect?

Biology
2 answers:
Ket [755]3 years ago
8 0

Answer:

fewer clouds

less precipitation

Explanation:

The amount of clouds in the sky is a strong indication of the level of pressure in a given area. This is because areas that are influenced by high pressure usually have few clouds and, at times, have no cloud at all and therefore have little or no probability of precipitation.

On the other hand, areas with low pressure influence present high cloudiness, with a strong presence of clouds. These areas are often called the cold front and have a high chance of precipitation.

leva [86]3 years ago
5 0
D. less precipitation

A high-pressure area, high or anticyclone is a region where the atmospheric pressure at the surface of the planet is greater than its surrounding environment. Winds within high-pressure areas flow outward from the higher pressure areas near their centers towards the lower pressure areas<span> further from their centers.</span>
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What is the probability of a couple having 5 daughters in a row? please help
worty [1.4K]

Answer:

3.125%

Explanation:

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2 years ago
assume that life on Mars requires cell potential to be 100mV, and the extracellular concentrations of the three major species ar
svetlana [45]

Answer:

Explanation:

From the information given:

The cell potential on mars E = + 100 mV

By using Goldman's equation:

E_m = \dfrac{RT}{zF}In \Big (\dfrac{P_K[K^+]_{out}+P_{Na}[Na^+]_{out}+P_{Cl}[Cl^-]_{out} }{P_K[K^+]_{in}+P_{Na}[Na^+]_{in}+ P_{Cl}[Cl^-]_{in}}      \Big )

Let's take a look at the impermeable cell with respect to two species;

and the two species be Na⁺ and Cl⁻

E_m = \dfrac{RT}{zF} In \dfrac{[K^+]_{out}}{[K^+]_{in}}

where;

z = ionic charge on the species = + 1

F = faraday constant

∴

100 \times 10^{-3} = \Big (\dfrac{8.314 \times 298}{1\times 96485} \Big) \mathtt{In}  \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\  53.57 = \dfrac{4}{[K^+]_{in}}

[K^+]_{in} = \dfrac{4}{53.57}

[K^+]_{in}  = 0.0476

For [Cl⁻]:

100 \times 10^{-3} = -0.0257 \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

-3.981 =  \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

0.01867 =  \dfrac{120}{[Cl^-]_{in}}

[Cl^-]_{in} = \dfrac{120}{0.01867}

[Cl^-]_{in} =6427.4

For [Na⁺]:

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

53.57= \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

[Na^+]_{in}= 2.70

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3 years ago
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