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2 years ago

Explain what takes place during the G2 phase of the cell cycle.

Chemistry

1 answer:

mrs_skeptik [129]2 years ago

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Answer:

The cell's DNA is replicated during the S stage as the first step to engaging in mitosis. Once DNA replication is complete, the cell enters the second gap stage, G2. During G2 the correct duplication of the DNA is verified and cell proteins necessary for cell division are produced.

Explanation:

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According to Bohr atomic model

nata0808 [166]

Answer:

A small positively charged nucleus surrounded by revolving negatively charged electrons in fixed orbits

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3 years ago

Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a

Inessa [10]

Different isotopes of the same element emit light at slightly different wavelengths, the minimum number of slits is mathematically given as

N=1820slits

<h3>What minimum number of slits is required to resolve these two wavelengths in second-order?</h3>

Generally, the equation for the wave is mathematically given as

$d\ sin\ (\theta\ m) \ = \ m\ \lambda$

Where the chromatic resolving power (R) is defined by

$R\ =\ \lambda\ / \ d \ \lambda$

R = nN,

Therefore

$\lambda_1 \ = \ (656.45)(1 \ * \ 10^{-9})/1mm$

$\lambda_1= 656.45*10^{-9}$

and

$\lambda_2= (656.27)(1*10^{-9})/1mm$

$\\\\\lambda_2= 656.27*10^{-9}m$

In conclusion, the minimum number of slits is required to resolve these two wavelengths in second-order

$N\ =\ \dfrac{\lambda}{m\ d\ T\ }\\\\$

Therefore

$N\ =\ \dfrac{656.45 \ * \ 10^{-9}}{2\ * \ (0.18*10^{-9})}$

N=1820slits

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1 year ago

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.3×1015Hz?

Sonja [21]

Answer:

We know that

ħf = ф + Ekmax

where

ħ = planks constant = 6.626x10^-34 J s

f = frequency of incident light = 1.3x10^15 /s (1 Hz =

1/s)

ф = work function of the cesium = 2.14 eV

Ekmax = max kinetic energy of the emmitted electron.

We distinguish that:

1 eV = 1.602x10^-19 J

So:

2.14 eV x (1.602x10^-19 J / 1 eV) = 3.428x10^-19 J

So,

Ekmax = (6.626x10^-34 J s) x (1.3x10^15 / s) - 3.428x10^-19 J

= 8.6138x10^-19 J - 3.428x10^-19 J = 5.1858x10^-19 J

Answer:

5.19x10^-19 J

Kinetic energy:

In physics, the kinetic energy of an object is the energy that it owns due to its motion. It is defined as the work required accelerating a body of a given mass from rest to its specified velocity. Having expanded this energy during its acceleration, the body upholds this kinetic energy lest its speed changes.

Answer details:

Subject: Chemistry

Level: College

Keywords:

• Energy

• Kinetic energy

• Kinetic energy of emitted electrons

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