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3 years ago

Use hormone and feedback mechanism in the same sentence

2 answers:

7 0

The answer to your question is : Hormones contact you through your body's system in order to receive a feedback from another point of view or perspective.

mylen [45]3 years ago

3 0

In order to gauge the feedback mechanism of the way your hormone responds, you need at least three hours to monitor the situation.

Hope that helped!!!

You might be interested in

There are two naturally occurring isotopes of bromine. 79Br has a mass of 78.9183 amu. 81Br has a mass of 80.9163 amu. Determine

Ede4ka [16]

Atomic mass of Br = 79.904
Now lets say y% is abundance of 79Br. 
Then abundance of 81Br = (100 - y) 
mass due to 79Br = 78.9183 * y/100 = 0.789183y
mass due to 81Br = 80.9163 x (100 - y)/100 = 0.809163(100 - y) 
Therfore
0.789183y+ 0.809163(100 - y) = 79.904 
0.789183y + 80.9163 - 0.809163y = 79.904 
 - 0.01998y= 79.904 - 80.9163
= - 1.0123 
y = 1.0123/0.01998 = 50.67% 
 79Br = 50.67% 
now
81Br = 100 - 50.67 = 49.33%
hope this helps

4 0

4 years ago

What happens when grapes get fermented

ValentinkaMS [17]

When grapes become fermented, they turn into wine.

6 0

3 years ago

Why does the body need nitrogen compounds?

Nostrana [21]

To make carbohydrates

4 0

3 years ago

Read 2 more answers

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to

Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density $\rho_{avg}$ ; we have:

$\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$

The average atomic weight is:

$A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$

So; in terms of vanadium, the Concentration of iron is:

$C_{Fe} = 100 - C_v$

From a unit cell volume $V_c$

$V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}$

where;

$N_A$ = number of Avogadro constant.

SO; replacing $V_c$ with $a^3$ ; $\rho_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$ ; $A_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$ and

$C_{Fe}$ with $100-C_v$

Then:

$a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }$

Replacing the values; we have:

$(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }$

$2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }$

$2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})$

$\mathbf{C_v = 9.1 \ wt\%}$

4 0

3 years ago

What makes an element inert?

ozzi

"The inert gases are obtained by fractional distillation of air, with the exception of helium which is separated from a few natural gas sources rich in this element, through cryogenic distillation or membrane separation. For specialized applications, purified inert gas shall be produced by specialized generators on-site. They are often used by chemical tankers and product carriers (smaller not a big as well as the tendency of inert gases vesselshtop specialized generators are also available for laboratories."

3 0

3 years ago