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elena-14-01-66 [18.8K]
3 years ago
6

Which type of irb review does not require an irb approval but does require a determination by the irb or an irb designee?

Physics
1 answer:
Westkost [7]3 years ago
4 0

Answer:

For this case the answer would be exampt. Since this is the only level that require a determination by the irb or an irb designee.

Explanation:

Previous concepts

We have 3 levels of the IRB review from the lowest to the highest:

Exempt. For thi slevel we have less than the minimal risk. And we have some examples like a survey with anonymous participants. And not classifed info. Other examples are: (Education research,  Surveys, interviews, educational tests, public observations (that do not involve children) ,Benign behavioral interventions ,Analysis of previously-collected, identifiable info/specimens ,Federal research/demonstration projects ,Taste and food evaluation studies).

Expedited. For this level we have more risk than th exempt level. An example would be analyze some info for non reasearch pruposes, but we can identify the subjects of the study. Other examples are :(Clinical studies of drugs and medical devices only when certain conditions are met , Collection of blood samples by finger stick, heel stick, ear stick, or venipuncture in certain populations and within certain amounts , Prospective collection of biological specimens for research purposes by noninvasive means , Research involving materials (data, documents, records, or specimens) that have been collected, or will be collected solely for non-research purposes , Collection of data from voice, video, digital, or image recordings made for research purposes )

Full board. That's the level with the maximum risk. Have included the identification of the subjects and the desire variable analyzed.

For our question:

Which type of irb review does not require an irb approval but does require a determination by the irb or an irb designee?

For this case the answer would be exampt. Since this is the only level that require a determination by the irb or an irb designee.

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Answer:

\frac{I}{I_0}=113.68

Explanation:

P = Acoustic power = 63 µW

r = Distance to the sound source = 210 m

Acoustic power

P=IA\\\Rightarrow I=\frac{P}{A}\\\Rightarrow I=\frac{63\times 10^{-6}}{4\times \pi \times 210^2}

Threshold intensity = I_0=1\times 10^{-12}\ W/m^2

Ratio

\frac{I}{I_0}=\frac{\frac{63\times 10^{-6}}{4\times \pi \times 210^2}}{1\times 10^{-12}}\\\Rightarrow \frac{I}{I_0}=113.68

Ratio of the acoustic intensity produced by the juvenile howler to the reference intensity is 113.68

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An air-track glider with a mass of 239 g is moving at 0.81 m/s on a 2.4 m long air track. It collides elastically with a 513 g g
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Answer:

Glider it stops just when it reaches the end of the runway

Explanation:

This is a shock between two bodies, so we must use the equations of conservation of the amount of movement, in the instant before the crash and the subsequent instant, with this we calculate the second glider speed, as the shock that elastic is also keep it kinetic energy

        Po = pf

        Ko = Kf

 Before crash

       Po = m1 Vo1 + 0

       Ko = ½ m1 Vo1²

 

After the crash

       Pf = m1 Vif + Vvf

       Kf = ½ m1 V1f² + ½ m2 V2f²

 

      m1 V1o = m1 V1f + m2 V2f           (1)

      m1 V1o² = m1 V1f² + m2 V2f²      (2)

We see that we have two equations with two unknowns, so the system is solvable,  we substitute in 1 and 2

   

     m1 (V1o -V1f) = m2 V2f      (3)

      m1 (V1o² - V1f²) = m2 V2f²

Let's use the relationship      (a + b) (a-b) = a² -b²

     m1 (V1o + V1f) (V1o -V1f) = m2 V2f²

We divide  with 3 and simplify

      (V1o + V1f) = V2f      (4)

Substitute in 3, group and clear

         m1 (V1o - V1f) = m2 (V1o + V1f)

         m1 V1o - m2 V1o = m2 V1f + m1 V1f

         V1f (m1 -m2) = V1o (m1 + m2)

         V1f = V1o  (m1-m2 / m1+m2)

We substitute in (4) and group

         V2f = V1o + (m1-m2 / m1 + m2) V1o

         V2f = V1o [1+ + (m1-m2 / m1 + m2)]

         V2f = V1o (2m1 / (m1+m2)

We calculate with the given values

         V1f = 0.81 (239-513 / 239 + 513)

         V1f = 0.81 (-274/752)

         V1f = - 0.295 m/s

The negative sign indicates that the planned one moves in the opposite direction to the initial one

         V2f = 0.81 [2 239 / (239 + 513)]

        V2f = 0.81 [0.636]

        V2f = 0.515 m / s

Now we analyze in the second glider movement only, we calculate the energy and since there is no friction,

         Eo = Ef

Where Eo is the mechanical energy at the lowest point and Ef is the mechanical energy at the highest point

         Eo = K = ½ m2 vf2²

         Ef = U = m2 g Y

   

         ½ m2 v2f² = m2 g Y

         Y = V2f² / 2g

         Y = 0.515²/2 9.8

         Y = 0.0147 m

At this height the planned stops, let's use trigonometry to find the height at the end of the track of the track

         tan θ = Y / x

         Y = x tan θ

The crash occurs in the middle of the track whereby x = 1.2 m

        Y = 1.2 tan 0.7

        Y = 0.147 m

As the two quantities are equal in glider it stops just when it reaches the end of the runway

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3 years ago
A 65 N boy sits on a sled weighing 52 N on a horizontal surface. The coefficient of friction between the sled and the snow is 0.
pogonyaev

Answer:

1.40 N

Explanation:

The magnitude of the frictional force is given by:

F=\mu N

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\mu is the coefficient of friction

N is the magnitude of the normal reaction

The coefficient of friction for this problem is \mu=0.012. The magnitude of the normal reaction is equal to the combined weight of the boy and the sled, because the surface is horizontal, so

N=65 N+52 N=117 N

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F=\mu N=(0.012)(117 N)=1.40 N

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