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4 years ago

What is the difference between the control group and the experimental group in an experimental study?

Physics

2 answers:

nikitadnepr [17]4 years ago

8 0

Answer:

Explanation:

The only difference between the control and experimental group is the independent variable.

Talja [164]4 years ago

7 0

The controller doesn’t change and experimental does

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HELP PLEASE NOWWWW !!!! ( 1 ) Jill and Scott both road their bikes for 30 minutes. Jill traveled 5 kilometers and Scott trave

katovenus [111]

Jill and Scott both moves for 30 minutes

now if Jill cover 5 km distance and Scott cover 10 km distance

now we know that the formula of speed is given as

$speed = \frac{distance}{time}$

now we will have

speed of Jill

$v = \frac{5}{0.5} = 10 km/h$

speed of Scott

$v = \frac{10}{0.5} = 20 km/h$

so correct answer here is

Scott had the faster speed since he rode at 20 k/h while Jill only traveled 10 km/h.

#2

distance travelled by each car is given as

$d = 400 miles$

now here it is given that

time taken by green car

$t_1 = 10 hours$

time taken by yellow car

$t_2 = 8 hours$

now we can find the speed of two cars

$speed = \frac{distance}{time}$

speed of green car

$v = \frac{400}{10} = 40 mph$

speed of yellow car

$v = \frac{400}{8} = 50 mph$

so correct answer will be

The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

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3 years ago

Read 2 more answers

How many millimeters are there in one centimeter

Anika [276]

The answer is 10 millimeters in one centimeter

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3 years ago

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A 5 cm radius isolated conducting sphere is charged so its potential is 100 V, relative to the potential far away. The charge de

MArishka [77]

The electric potential (V) produced by a point charge with a charge of magnitude Q, at a point a distance r away from the point charge, is given by the equation:

$V = \frac{kQ}{r}$

Here,

k = Coulomb's constant

Q = Charge

r = Distance

If we rearrange the equation to find the distance we have,

$Q=\frac{Vr}{k}$

$Q = \frac{(100)(5*10^{-2})}{9*10^9}$

$Q = 5.55*10^{-10} C$

Now the value of the charge density is equivalent to the charge on the surface area of the sphere this is

$\gamma = \frac{Q}{A_s}$

$\gamma = \frac{Q}{4\pi r^2}$

$\gamma = \frac{ 5.55*10^{-10}}{4\pi (5*10^{-2})^2}$

$\gamma = 1.76*10^{-8} C/m^2$

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3 years ago

How can i pass physics exam in 3 days?

alexira [117]

Explanation:

you can pass the exam in 3 days by doing labouring hard, by understanding and by thinking theoretical and practising practically.

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3 years ago

How does the speed of the wave compare between 300 hz and 400 hz

scoray [572]

The ratio between initial and final speed is 3:4

Explanation:

The speed of a wave is given by the wave equation:

$v=f\lambda$

where

v is the speed

f is the frequency of the wave

$\lambda$ is the wavelength

In this problem, we have a wave with wavelength $\lambda$ and initial frequency

$f_1 = 300 Hz$

So its speed can be written as

$v_1 = f_1 \lambda = 300 \lambda$

Later, its frequency changes to

$f_2 = 400 Hz$

Assuming that its wavelength has not changed, this means that its new speed is

$v_2 = f_2 \lambda = 400 \lambda$

By calculating the ratio between the two,

$\frac{v_1}{v_2}=\frac{3}{4}$

So, the ratio between initial and final speed is 3:4.

Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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4 years ago

What is the difference between the control group and the experimental group in an experimental study?​

What is the difference between the control group and the experimental group in an experimental study?