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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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sergiy2304 [10]

Answer:

B, A, C

Explanation:

3 0
3 years ago
On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of
Advocard [28]

Answer:

On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of electrons? (<em>the mass will decrease by a very small factor</em>)

(b) What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons?  (<em>The mass will increase by a very small factor</em>)

Explanation:

(a) On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of electrons? (<em>the mass will decrease by a very small factor</em>)

The mass of an atom is given by the sum of the masses of the protons, neutrons and electrons. Electrons has lower mass than protons and neutrons, so they have a minor contribution to the total mass of the atom.    

When an object is electrically neutral it means that it has the same number of protons and electrons. For the case of an object positively charged, the rate of protons is greater than the number of electrons. That means that atom lose electrons so the mass will decrease in a very small factor.

(b) What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons?  (<em>The mass will increase by a very small factor</em>)

For the case when the object is negatively charged, it means that the atom gains electrons from another object, leading to the conclusion that the mass of the atom will increase in a very small factor.  

Key values:

Electron mass: 9.1095×10⁻³¹ Kg

Proton mass: 1.67261×10⁻²⁷ Kg

Neutron mass: 1.67492×10⁻²⁷ Kg

5 0
3 years ago
An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t
mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a)  the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F_L = C_L\frac{1}{2}ρV²A = W

we substitute

0.45 × \frac{1}{2} × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F_D = T = C_D\frac{1}{2}ρV²A

we substitute

= 0.065 × \frac{1}{2} × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0
3 years ago
A 65 kg student sits 3 m away from a 70 kg student. What is the magnitude of the gravitational force between the two students?
LiRa [457]

Answer:

F=3.37\times 10^{-8}\ N

Explanation:

Given that,

Mass of student 1, m₁ = 65 kg

Mass of student 2, m₂ = 70 kg

The distance between the students, d = 3 m

We need to find the magnitude of the gravitational force between the two students. The formula for the magnitude of the gravitational force between two masses is given by :

F=G\dfrac{m_1m_2}{d^2}\\\\F=6.67\times 10^{-11}\times \dfrac{65\times 70}{(3)^2}\\F=3.37\times 10^{-8}\ N

So, the gravitational force between the two students is 3.37\times 10^{-8}\ N.

3 0
2 years ago
A wheel rotates clockwise 10 times per second. what is its angular speed answer
bogdanovich [222]
The frequency of the wheel is the number of revolutions per second:
f= \frac{N_{rev}}{t}= \frac{10}{1 s}=10 Hz
And now we can calculate the angular speed, which is given by:
\omega = 2 \pi f=2 \pi (10 Hz)=62.8 rad/s in the clockwise direction.
3 0
3 years ago
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