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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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A spring has an unstretched length of
koban [17]

Answer:

24 N

Explanation:

Given,

Unstretched length or original length , x1 = 12cm

Stiffness constant, k = 8 N/cm

Load required to take spring to a length , x2 of 15cm

Recall the relation :

F = Ke

Where, e = extension

e = x2 - x1

e = (15 - 12) = 3

F = ke

F = 8 N/cm * 3cm

F = 24 N/cm*cm

F = 24 N

Hence, required load = 24 N

8 0
3 years ago
A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it
asambeis [7]

Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

4 0
3 years ago
1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is i
ad-work [718]

Answer: 2.67 m/s

Explanation:

Given

Mass of block A  is m_a=2\ kg

mass of block B is m_b=3\ kg

The initial velocity of block A u_a=5\ m/s

the initial velocity of block B is u_b=0

After collision velocity of block A is v_a=1\ m/s

Conserving momentum

m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s

The momentum of block A after the collision is P_a=2\times 1=2\ kg.m/s

Therefore, there is no change in sign.

6 0
3 years ago
HELP PLEASE<br> (Look at the picture)
kotykmax [81]

Answer:

what is that?

Explanation:

i dont know what us that sorry

3 0
3 years ago
1. An electron travels 4.82 meters in 0.00360 seconds. What is its average speed?
vredina [299]

Answer:

speed =distance /time

speed =4.82/0.00360

speed =1338.8m/s

6 0
3 years ago
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