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polet [3.4K]
3 years ago
10

Megan wants to see if mice that are fed food A or B grow at a faster rate. She measures the mass of the mice on day 1, then divi

des them into two groups based on the type of food they will be given. She feeds each group the same amount of food at the same time each day. Does Megan have a control group in this experiment?
Physics
2 answers:
diamong [38]3 years ago
7 0
<span>No, there is no control group because each group is treated under test conditions.</span>
Nat2105 [25]3 years ago
4 0

Answer:

No

Explanation:

It is an example of experimental group. Experimental group is defined as the group in an experiment under which each variable is tested, however, one variable is tested at a time.

On the other hand, control group is defined as the group which does not receive treatment by the researchers.

The only difference between the both group is that, under experimental group independent variable is changed but under control group it is held constant.

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AN airplane travels 4000 m in 20s on a heading 0f 35 degrees north west. Calculate average velocity .
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I can not find the answer to the one after potassium ?
Zepler [3.9K]

Answer:

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3 0
3 years ago
Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of
Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = \frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = \sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}

(|r₂₁|)² = 148.25

F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }

      = 0.050938(0.19107i + 0.54933j) N

      = (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0
3 years ago
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