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3 years ago

What does a control group show in an experimental investigation?

2 answers:

8 0

The control group, receiving no intervention, is used as a baseline to compare groups and assess the effect of that intervention.

jonny [76]3 years ago

7 0

The control group in an experimental investigation shows how nature would act without manipulation by the researchers.

Answer: Option B

<u>Explanation: </u>

The control group, a separated group from the rest of all other scientific experiments. The work done by this group does not affect the results of the actual experiment. The significant work of this group is to analyse the independent variables and their influence on the experiment. The control group is also classified as positive group and negative group.

The work of positive group is to check whether the independent variables were affecting the other parameters in a positive way to get the desired results. While the negative group composes the constraints to be followed by the investigators to get the desired results. Thus the control group shows how nature would act without manipulation by the researchers.

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NemiM [27]

Answer:

what is your question because I seem to not see a question in this question....

3 0

3 years ago

A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red

makkiz [27]

Answer:

Explanation:

$\lambda$ = Observed wavelength = 550 nm

$\lambda'$ = Actual wavelength = 635 nm

c = Speed of light = $3\times 10^8\ \text{m/s}$

v = Velocity of the physicist

Doppler shift is given by

$f=\sqrt{\dfrac{c+v}{c-v}}f'\\\Rightarrow \dfrac{c}{\lambda}=\sqrt{\dfrac{c+v}{c-v}}\dfrac{c}{\lambda'}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}=\dfrac{c+v}{c-v}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}=\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}(1-\dfrac{v}{c})=1+\dfrac{v}{c}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}(1-\dfrac{v}{c})=1+\dfrac{v}{c}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}-1=\dfrac{v}{c}(1+\dfrac{\lambda'^2}{\lambda^2})\\\Rightarrow v=\dfrac{c(\dfrac{\lambda'^2}{\lambda^2}-1)}{1+\dfrac{\lambda'^2}{\lambda^2}}$

$\Rightarrow v=\dfrac{3\times 10^8\times (\dfrac{635^2}{550^2}-1)}{1+\dfrac{635^2}{550^2}}\\\Rightarrow v=42817669.77\ \text{m/s}$

The physicist was traveling at a velocity of $42817669.77\ \text{m/s}$ .

4 0

3 years ago

a mass is measured as 6.2kg in one experiment and as 6.20kg in another. in what way do these two measurements differ?

Lubov Fominskaja [6]

Answer:

They do not differ because 6.2 is the same as 6.20

Explanation:

8 0

3 years ago

Read 2 more answers

Question #14

Katena32 [7]

The decrease in energy in the hydrogen molecule is what allows its formation on Earth, but in stars the great energy of the explosion has a kinetic energy so great that electrons cannot bind to another atom, which is why hydrogen has a single atom.

The hydrogen molecule is a form that two hydrogen atoms share their electrons decreasing the total energy of the molecule, this bond has a covalent and hydrogen bonding characteristic.

In a stellar explosion, the energy released increases the energy of the hydrogen atom, for which we have two possibilities:

Its electron is lost, so we are in a single proton, in the case of structures where the proton and the elector are

The hydrogen atom remains but the energy of the atom is very high so the kinetic energy of the electron prevents the electron from being shared by the other atom and the molecule cannot be formed.

When the atoms are thrown into space, the separation between them is so high that it does not allow electrons to be shared and molecules cannot be formed either.

In conclusion, the decrease in energy in the hydrogen molecule is what allows its formation on Earth, but in stars the great energy of the explosion has a kinetic energy so great that electrons cannot join another atom, which is why the hydrogen has only one atom.

Learn more about the Hydrogen atom here:

brainly.com/question/22464200

6 0

3 years ago

A heat engine exhausts 8900 j of heat while performing 2800 j of useful work. part a what is the efficiency of this engine?

Anuta_ua [19.1K]

Given: Heat Qout means useful work = 2800 J

Heat Qin = 8900 J

Required; Efficiency = ?

Formula: Efficiency = Qout/Qin = x 100%

= 2800 J/8900 J = 0.3146 X 100 %

Efficiency = 31.46%

7 0

3 years ago