All categories

3 years ago

Which is the fastest moving component in the electropherogram?

2 answers:

6 0

An electropherogram, or electrophoregram, is a record or chart produced when electrophoresis is used in an analytical technique, primarily in the fields of molecular biology or biochemistry.

...

These genotypes can be used for:

genealogical DNA testing.

DNA paternity testing.

DNA profiling.

phylogenetics.

population genetics

hope this helped :)

Inessa [10]3 years ago

4 0

Answer:

Explanation:

I had this question and y was right .

You might be interested in

What is torricellian vaccum<br>

motikmotik

Answer:

a vacuum formed above the column of mercury in a mercury barometer.

6 0

3 years ago

Read 2 more answers

Two cars are moving in the same direction at the same speed of (72.0 kmh). What is

9966 [12]

Answer:0

Explanation: at same direction

72-72=0

3 0

3 years ago

How is energy can be changed from one state to another?

snow_lady [41]

Answer:

Explanation:

Heat is probably the easiest energy you can use to change your physical state. The atoms in a liquid have more energy than the atoms in a solid. There is a special temperature for every substance called the melting point.They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. EXAMPLE: Solid ice melting into liquid

5 0

4 years ago

The induced magnetic field at radial distance 4.0 mm from the central axis of a circular parallel-plate capacitor is 1.8 ✕ 10−7

wel

Answer:

$\frac{dE}{dt}=2.07*10^{13}\frac{V/m}{s}$

Explanation:

According to Gauss's law, the electric flux through the circular plates is defined as the electric field multiplied by its area:

$\Phi=EA=E(\pi R^2)(1)$

The magnetic field around the varying electric field of the circular plates is given by:

$B=\frac{\epsilon_0 \mu_o}{2\pi r}\frac{d\Phi}{dt}(2)$

Replacing (1) in (2) and solving for $\frac{dE}{dt}$ :

$B=\frac{\epsilon_0 \mu_o\pi R^2}{2\pi r}\frac{dE}{dt}\\\frac{dE}{dt}=\frac{2rB}{\epsilon_0 \mu_o R^2}\\\frac{dE}{dt}=\frac{2(4*10^{-3}m)(1.8*10^{-7}T)}{(8.85*10^{-12}\frac{C^2}{N\cdot m^2})(4\pi *10{-7}\frac{Tm}{A})(2.5*10^{-3}m)^2}\\\\\frac{dE}{dt}=2.07*10^{13}\frac{V/m}{s}$

3 0

3 years ago

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$