How is the control group and experimental group different

1 answer:

4 0

The control group is the independent variable and the experimental group is the dependent due to change during the experiment. The experimental group will usually rely on another variable in the experiment for change.

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Darren filled ocean water, fresh water, bottled water, and tap water into four different containers. He then dropped identical g

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I would say container 1

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Carbon monoxide pure substance or mixture and why

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Carbon monoxide is a mixture cause it's a mixture of
different chemical compounds

such as air and CO but they're chemically mixed together which makes it no longer pure

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What is the acceleration of a boy on a skateboard if the net force on the boy is 15 N? The total mass of the boy and the skatebo

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   Force = (mass) x (acceleration)

If the full 15N is pointing parallel to the ground,
then

   15 N = (58 kg) x (acceleration).

Divide each side
by 58 kg: Acceleration = 15 N / 58 kg

   = (15 kg-m/s²) / (58 kg)

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Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m

Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

$F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\$

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

$F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC$

In words, the value of q₃ must be 5.3nC.

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