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myrzilka [38]
3 years ago
8

A researcher is interested in exploring the effects of exposure to ultraviolet light on people’s emotional state. He divides his

participants into three groups. The first group receives 40 minutes of ultraviolet light 3 times a week, the second group receives 20 minutes of ultraviolet light 3 times a week, and the third group receives no ultraviolet light. Tell me which group(s) were in the control group, and which were in the experimental group. (You can simply delete the incorrect answer for each group. So, for example, if you think the answer to group one is "control," you can simply delete the word "experimental" after group one)
Group One: Experimental Control

Group Two: Experimental Control

Group Three: Experimental Control
Physics
1 answer:
bekas [8.4K]3 years ago
4 0

Answer:

Explanation:

In an experimental research, the control group is the group that serves as the neutral group that is not given any form of treatment and serves as the group in which the experimental groups are firstly compared to. Thus, <u>the control group in the question described is the Third group</u>.

While experimental groups are the groups that receive treatments required to make an inference from the experiment. From this description, <u>it can be deduced that the First and the Second group are the experimental groups.</u>

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ASAP do all will give 25 per person pls will mark brainiest
neonofarm [45]

Answer:

1) true

2) false

3) false

4) true

5) true

6) true

7) true

8) false

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i think these are correct if im wrong on a few im sorry. Hope this helps at least a bit. And if i do get some wrong you know just to pick the opposite answer.

4 0
3 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

5 0
3 years ago
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Lelu [443]
Surface air pressure is a consequence of the weight of the air acting on its surface. For example, if you are standing on Mars, the pressure around you is what you call the surface air pressure. Thus, that surface air pressure must be 0.007 atm.
5 0
3 years ago
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Answer:

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Explanation:

p=mv

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skateboard b

p=(50kg)(2m/s)=100kg*m/s

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3 years ago
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MariettaO [177]
The correct answer is A
4 0
3 years ago
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