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Sonbull [250]
3 years ago
13

A circle with circumference \blue{8}8start color #6495ed, 8, end color #6495ed has an arc with a 288^\circ288

Biology
1 answer:
vivado [14]3 years ago
3 0

Answer:

The arc is 6.4 units long.

Explanation:

For a circle of radius R, the circumference is given by:

C = 2*pi*R

If we have a section of this circle defined by an angle θ (such that this section makes an arc), the length of that arc is:

A = (θ/360°)*2*pi*R

and 2*pi*R is the circumference, then we can write the length of the arc as:

A = (θ/360°)*C

qsIn this case, we know that the circumference is equal to 8 units, and the arc has an angle of 288°

Then the length of that arc is:

A = (288°/360°)*C = (288°/360°)*8 = 6.4

The arc is 6.4 units long.

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Answer:

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Explanation:

Function:

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When skin cells have third-degree burns, skin grafts are used. What is the benefit to using skin grafts?
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Identify the lysogenic virus: A. Flu B. Smallpox C. Measles D. Chickenpox
alexandr1967 [171]

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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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The answer is Speciation cannot occur in the absence of mutation. Speciation cannot occur in the absence of natural selection

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