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3 years ago

N which parts of the cell cycle would chromosomes be paired up? Check all that apply.

Chemistry

1 answer:

Kitty [74]3 years ago

4 0

Answer:

The correct option es interphase

Explanation:

The interface is a stage of the cell division where a lot of energy is demanded, and it is the longest of the entire cycle.

This is because it is here where the duplication of the genetic load is generated, that is, of the chromosomes.

The chromosomes reach the cell poles and this is how a contraction begins to appear in the middle of the cell, it is essential to clarify that the interface not only requires energy but also occupies 95% of the cell division.

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Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank. The outlet volumetric flow

Montano1993 [528]

Explanation:

Apply the mass of balance as follows.

Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank

$\frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)$

$\rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h$

$\frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}$

[/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = $\frac{10}{\rho A_{c}}$

$A_{c} = 0.01 m^{2}$

$\frac{dh}{dt}$ + h = 1

$\frac{dh}{dt}$ = 1 - h

$\frac{dh}{1 - h}$ = dt

$\frac{ln(1 - h)}{-1}$ = t + C

Given at t = 0 and V = 0

$A \times h$ = 0

or, h = 0

-ln(1 - h) = t + C

Initial condition is -ln(1) = 0 + C

C = 0

So, -ln(1 - h) = t

or, t = $ln (\frac{1}{1 - h})$ ........... (1)

(a) Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.

t = $ln (\frac{1}{1 - h})$

t = $ln (\frac{1}{1 - 0.6})$

= $ln (\frac{1}{0.4})$

= 0.916 seconds

(b) As maximum height of water level in the tank is achieved at steady state that is, t = $\infty$ .

1 - h = exp (-t)

1 - h = 0

h = 1

Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.

8 0

3 years ago

If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperature of 1100˚C for 8 hours, wh

RideAnS [48]

Answer:

8.354 nanometers

Explanation:

To treat a diffusive process in function of time and distance we need to solve 2nd Ficks Law. This a partial differential equation, with certain condition the solution looks like this:

$\frac{C_{s}-C{x}}{C_{s}-C_{o}}=erf(x/2\sqrt{D*t})$