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mihalych1998 [28]

4 years ago

14

3. The nervous system works together to communicate with different parts of the body and with the external environment. Spinal c

ord damage can affect the whole system because the spinal cord __________. A. controls the actions of the brain and peripheral nerves B. is the most important organ in the nervous system C. mediates communication from the brain to the rest of the body D. is primarily involved in communication with the external environment

2 answers:

Scorpion4ik [409]4 years ago

8 0

I don’t know I just need points

Likurg_2 [28]4 years ago

7 0

Answer:

C. mediates communication from the brain to the rest of the body

Explanation:

The brain and spinal cord are your body's central nervous system. The brain is the command center for your body, and the spinal cord is the pathway for messages sent by the brain to the body and from the body to the brain.

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Which one of the following statements does not accurately describe vibrations?
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yKpoI14uk [10]

The height increases (c)

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3 years ago

The apparent backward motion of a planet along the celestial sphere is called ____________ motion.

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Explanation:

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3 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

3 years ago