HEY CAN YALL HELP ME IN DIS!!

Which of the following is the best reason for using a scanning electron microscope?

torisob [31]

Answer is C: Ability to see three-dimensional images of the surfaces of object

Explanation:

To enable the technician see fractures and broken particles in a better resolution as the SEM sees the peaks and valley of the structure.

8 0

3 years ago

What is potential energy?

kvasek [131]

Answer:

Explanation:Here's li $^{}$ nk to the answerly/3fcEdSx:

bit. $^{}$

4 0

3 years ago

How fast (in rpm) must a centrifuge rotate if a particle 6.00 cm from the axis of rotation is to experience an acceleration of 1

astra-53 [7]

The acceleration experienced by the particle is given by
$a=113000 g=113000 \cdot 9.81 m/s^2$
This corresponds to the centripetal acceleration of the motion, which is related to the angular speed $\omega$ of the particle and its distance r from the axis by the relationship
$a= \omega ^2 r$
In our problem, $r=6 cm=0.06 m$ , so we can solve for $\omega$ :
$\omega = \sqrt{ \frac{a}{r} } = \sqrt{ \frac{113000 \cdot 9.81 m/s^2}{0.06 m} }=4298 rad/s$
However, we must convert it into rpm (revolution per minute).
We know that 1 rad corresponds to $( \frac{1}{2 \pi} )$ revolutions, while $1 s = \frac{1}{60} min$ . So we the conversion is $\omega = 4298 rad/s \cdot ( \frac{1}{2\pi} rev/rad )( 60 s/min)=41067 rpm$

4 0

4 years ago

PLEASE I NEED HELP CLICK ON THIS IMAGE

3241004551 [841]

I believe you are incorrect. A weathered mountain would appear more jagged.

I do believe with a lot of exposure to weather will make the mountain appear somewhat more jagged compared to a mountain that is less weathered.

<em>If this is incorrect, please, don't refrain to tell me.</em>

6 0

3 years ago

Helium gas is in a cylinder that has rigid walls. If the pressure of the gas is 2.00 atm, then the root-mean-square speed of the

aivan3 [116]

Answer:

$p=p_{2}-p_{1}=2.9188atm$

Explanation:

First step

To calculate by how much the pressure has been increased we must find a relation between Vrms and the pressure p

$V_{rms}=\sqrt{\frac{3RT}{M} } \\$

Where M is molar mass,R is gas constant and T is temperature

As R and M is constant so we obtain

Vrams∝√T

Second step

From Ideal gas law we know that

$pV=nRT\\$

we will find that

p∝T

So from first and second step, we can obtain that

Vrms∝√p

And a relation between both of them could be given by:

$\frac{V_{rms1} }{V_{rms2}}=\frac{\sqrt{p_{1} } }{\sqrt{p_{2}} }\\ p_{2}=[\frac{(V_{rms1})^{2} }{(V_{rms2})^{2} }]p_{1}\\ p_{2}=\frac{(276m/s)^{2} }{(176m/s)^{2} } (2atm)\\p_{2}=4.9188atm$

The pressure is increased to 4.1988 atm, so the amount will be given by:

$p=p_{2}-p_{1}=4.9188atm-2atm\\p=p_{2}-p_{1}=2.9188atm$

7 0

3 years ago