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Zepler [3.9K]
3 years ago
14

Valerie is preparing a blood sample for analysis. She wants to separate the solid red blood cells and obtain only the liquid por

tion, called plasma. The red blood cells are too small to filter out, but are denser than the plasma.
Describe the steps Valerie could take to separate the blood sample.
Physics
2 answers:
Citrus2011 [14]3 years ago
6 0

Sample Response: First, she should put the sample in a test tube and place it in a centrifuge. This would cause the red blood cells to move to the bottom because of their higher density. Next, she would be able to decant the plasma and analyze it separately from the red blood cells.

Hope this helps :)

liraira [26]3 years ago
4 0
<span>First, she should put the sample in a test tube and place it in a centrifuge. This would cause the red blood cells to move to the bottom because of their higher density. Next, she would be able to decant the plasma and analyze it separately from the red blood cells.</span>
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A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

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we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
3 years ago
A mass hanging from a spring is set in motion and its ensuing velocity is given by v (t )equals 2 pi cosine pi t for tgreater th
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Answer:

2(maximum), -2(minimum), -2(maximum).

Explanation:

V(t)= 2πcos πt--------------------------------------------------------------------------------(1).

Therefore, there is a need to integrate v(t) to get S(t).

S(t)= 2×sinπt + C ------------------------------------------------------------------------------(2).

Applying the condition given, we have s(0)= 0.

S(0)= 2sin ×π(0) + C.

Which means that; 0+C= 0. That is; C=0.

S(t)= 2 sin πt.

The mass moves to its highest positions at time,t=half(1/2=.5) and time,t=2.5.

Take note that;  sin(π/2) = sin(5π/2) = 1 .

Also, the mass moves to its lowest position at time,t=(3/2); also, sin(3π/2) = -1.

Therefore, we have that 2 maximum; -2 minimum and -2 maximum.

7 0
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Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message e
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Answer:

The duration  is  T  =72 \  years /tex]Explanation:From the question we are told that     The  distance is  [tex]D  =  35 \ light-years = 35 *  9.46 *10^{15} = 3.311 *10^{17} \  m

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converting to years

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Now the total time taken is mathematically represented as

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