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Bezzdna [24]
3 years ago
8

The brain mass of a human fetus during a particular trimester can be accurately estimated from the circumference of the head by

​m(c)equalsStartFraction c cubed Over 100 EndFraction minus StartFraction 1500 Over c EndFraction ​, where​ m(c) is the mass of the brain​ (in grams) and c is the circumference​ (in centimeters) of the head. a. Estimate the brain mass of a fetus that has a head circumference of 20 cm. b. Find the rate of change of the brain mass for a fetus that has a head circumference of 20 cm and interpret your result.
Physics
1 answer:
Paladinen [302]3 years ago
3 0

Answer:

a. If c = 20 cm, then the mass of the brain is m = 5 g.

b. At c = 20 cm, the brain's mass is increasing at a rate of 15.75 g/cm.

Explanation:

From the equation

m\left(c\right) = \frac{c^3}{100}-\frac{1500}{c}

we have

a. for c = 20 cm

m\left(20\right)=\frac{20^3}{100}-\frac{1500}{20}=5,

then the mass is m(20) = 5 g.

b. In order to find the rate of change, first we derivate

\frac{dm}{dc}=\frac{3c^2}{100}+\frac{1500}{c^2}.

Evaluated at c = 20 cm, we have

\frac{dm}{dc}|_{c=20}=\frac{3\times 20^2}{100}+\frac{1500}{20^2}=15.75.

So, at c = 20 cm, the mass of the brain is increasing at a rate of 15.75 g/cm.

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Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, \rho = 10.49 g/cm^{3}

Density of Pd, \rho = 12.02 g/cm^{3}

Atomic weight of Ag, A = 107.87 g/mol

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Now,

The average density, \rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}

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n = 4 = no. of atoms in FCC

Therefore,

\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}

Therefore, the length of the unit cell is given as:

a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}            (1)

Average atomic weight is given as:

A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}

where

C_{Ag} = 79 %

A_{Ag} = 107

C_{Pd} = 21%

A_{Pd} = 106

Therefore,

A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78

In the similar way, average density is given as:

\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}

\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}

Therefore, edge length is given by eqn (1) as:

a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm

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