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tatyana61 [14]
3 years ago
11

Researchers conducted an experiment to identify the effects of three different hand sanitizers on the growth of bacteria. The re

searchers grew one type of bacteria in twenty-four identical covered petri dishes. Then, they opened eighteen petri dishes and applied the three hand sanitizers to six petri dishes each. They left the remaining six petri dishes closed as a control group. When the researchers measured the growth of bacteria two days later, the six dishes that had no hand sanitizer showed the least amount of bacterial growth. The researchers concluded that not using any hand sanitizer would lead to the least bacterial growth.
Which statement best explains why the researchers’ conclusion was not justified?
A. The experiment tested too many variables.
B. The control group was not treated the same as the experimental group.
C. The sample size of twenty-four petri dishes is not large enough.
D. The single type of bacteria is not adequate to make any conclusions.
Physics
1 answer:
jek_recluse [69]3 years ago
5 0
The best and most correct answer among the choices provided by your question is the second choice or letter B.

The researchers’ conclusion was not justified because t<span>he control group was not treated the same as the experimental group.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o
polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges q_1 and q_2 separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

\displaystyle F=\frac{k\ q_1\ q_2}{r^2}

Where k is the proportional constant of value

k=9*10^9\ N.m^2/c^2

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

\displaystyle F_1=\frac{k\ Q\ Q}{d^2}

\displaystyle F_1=\frac{k\ Q^2}{d^2}

The force between charges separated 2d is

\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}

\displaystyle F_2=\frac{k\ Q^2}{4d^2}

And the force between the charges A and D is

\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}

\displaystyle F_3=\frac{k\ Q^2}{9d^2}

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})

\displaystyle F_A=-\frac{41}{36}F_1

Charge B. It receives force to the right from A and D and to the left from C

\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}

\displaystyle F_B=\frac{1}{4}F_1

Charge C. It receives forces to the right from all charges.

\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}

\displaystyle F_C=\frac{9}{4}F_1

Charge D. It receives forces to the left from all charges

\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}

\displaystyle F_D=-\frac{49}{36}F_1

Comparing the magnitudes of each force is just a matter of computing the fractions

\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36

a.

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

b.

The ratio of the greatest to the smallest net force is

\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9

The greatest force is 9 times the smallest net force

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