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MrRa [10]
3 years ago
14

What is the control group

Physics
2 answers:
otez555 [7]3 years ago
8 0

A control group in a scientific experiment is a group separated from the rest of the experiment, where the independent variable being tested cannot influence the results. This isolates the independent variable 's effects on the experiment and can help rule out alternative explanations of the experimental results.

skelet666 [1.2K]3 years ago
5 0

Answer:

members of a control group receive a standard treatment, a placebo, or no treatment at all.

Explanation:

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A 4 kg and 6 kg bowling ball are dropped from the same height at the same time. The two balls strike the ground at the same time
likoan [24]

It's actually B - The 6 kg ball has a greater force of gravity exerted on them


7 0
4 years ago
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A circuit with a 60 V battery, a 7 Ω resistor, a resistor with 0.2 A across it, and another 7 Ω resistor in series. What is the
skelet666 [1.2K]

Answer:

28.6 ohm

Explanation:

there is a 3rd resistor in series. voltage drop across the resistors will be equal to 60v

7*.2+7*.2 +.2x = 60

2.8+.2x = 60

.2x = 57.2

x = 28.6

8 0
2 years ago
Name two things radio waves have in common with visible light
kondaur [170]
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8 0
3 years ago
Which speed(s) is/are below the sound barrier in air? A. 257 m/s B. 1,030 m/s C. 10 m/s
Assoli18 [71]

The answers are A&C. please mark me brainliest

8 0
3 years ago
A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be cr
Grace [21]

Answer: 0.313 rad/s

Explanation:

The equation that relates the velocity V and the angular velocity \omega in the uniform circular motion is:

V=\omega.r   (1)

Where r=d/2=100m is the radius of the space station (with a diaeter of 200m) that describes the uniform circular motion.

Isolating \omega from (1):

\omega=\frac{V}{r}  (2)

On the other hand, we are told the “artificial gravity” produced by the cetripetal acceleration a_{c} is 9.8m/s^{2}, and is given by the following equation:

a_{c}=\frac{V^{2}}{r}   (3)

Isolating V:

V=\sqrt{a_{c}.r}   (4)

V=31.3049m/s   (5)

Substitutinng (5) in (2):

\omega=\frac{31.3049m/s}{100m}  (6)

\omega=0.313rad/s This is the angular velocity that would produce an “artificial gravity” of 9 9.8m/s^{2}.

6 0
3 years ago
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