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ad-work [718]
3 years ago
5

The sympathetic division of the autonomic nervous system prepares the body for stressful situations that require energy expendit

ure, such as by increasing heartbeat and respiratory rate to flee from a threatening situation.
A. True.
B. False.
Biology
1 answer:
k0ka [10]3 years ago
7 0

Answer:

True

Explanation:

The sympathetic autonomic nervous system activates the body for hormonal or neuronal responses also known as fight- or - flight response, For example during an fire emergency and there is a need to run. The response occurs primarily by  via impulses transmitted through the sympathetic nervous system, and also secondarily  through catecholamines secreted from the adrenal medulla.

Although the sympathetic autonomic nervous system is activated in stressful conditions, it needs to be constantly active even at a basal level to maintain homeostasis.

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Explain the 1st and 2nd law of thermodynamics in terms of food chain, energy flow, pyramid of number and pyramid of energy
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3 years ago
What is a reducing sugar?
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3 years ago
While alive, an organism absorbs (radioactive) carbon 14 at such a rate that the proportion of carbon 14 in the organism remains
Inga [223]

Answer:

The person has been dead for approximately 15,300 years

Explanation:

<u>Available data</u>:

  • The half-life of carbon 14 is 5,600 years
  • The human skeleton level of carbon 14 is 15% that of a living human

To answer this question we can make use of the following equation

Ln (C14T₁/C14 T₀) = - λ T₁

Where,

  • C14 T₀ ⇒ Amount of carbon in a living body at time 0 = 100%
  • C14T₁ ⇒ Amount of carbon in the dead body at time 1 = 15%
  • λ ⇒ radioactive decay constant = (Ln2)/T₀,₅
  • T₀,₅ ⇒ The half-life of carbon 14 = 5600 years
  • T₀ = 0
  • T₁ = ???

Let us first calculate the radioactive decay constant.

λ = (Ln2)/T₀,₅

λ = 0.693/5600

λ = 0.000123

Now, let us calculate the first term in the equation

Ln (C14T₁/C14 T₀) = Ln (15%/100%) = Ln 0.15 = - 1.89

Finally, let us replace the terms, clear the equation, and calculate the value of T₁.

Ln (C14T₁/C14 T₀) = - λ T₁

- 1.89 = - 0.000123 x T₁

T₁ = - 1.89 / - 0.000123

T₁ = 15,365 years

The person has been dead for approximately 15,300 years

4 0
3 years ago
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