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3 years ago

The treatment group in an experiment is also called the control group. Please select the best answer from the choices provided

Physics

2 answers:

BartSMP [9]3 years ago

8 0

It would be false it is not called the control group because the control group is something different

JulsSmile [24]3 years ago

5 0

The correct answer is:

false

The explanation:

It is not called a control group as the control group is a group of subjects closely resembling the treatment group in many demographic variables but not receiving the active medication or factor under study and thereby serving as a comparison group when treatment results are evaluated.

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The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed of 7,750 m/s, how long is HS

deff fn [24]

Answer: 5,640 s (94 minutes)

Explanation:

the tangential speed of the HST is given by

$v=\frac{2\pi r}{T}$ (1)

where

$2\pi r$ is the length of the orbit

r is the radius of the orbit

T is the orbital period

In our problem, we know the tangential speed: $v=7,750 m/s$ . The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

$r=6.38\cdot 10^6 m+569,000 m=6.95\cdot 10^6 m$

So, we can re-arrange equation (1) to find the orbital period:

$T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s$

Dividing by 60, we get that this time corresponds to 94 minutes.

6 0

3 years ago

Read 2 more answers

Calculate the average times it took the car to travel 0. 25 and 0. 50 meters. Record the averages, to two decimal places, in Tab

Illusion [34]

The average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

<h3>How to calculate the Average speed?</h3>

The average speed can be calculated by adding the speed of each trial divided by the number of trials,

For 0.25 m the average speed will be:

$S_{avg} = \dfrac{2.24 + 2.21 + 2.23}{ 3}\\\\S_{avg} = 2.22$

For the 0.50 m, the average speed will:

$S_{avg} = \dfrac {3.16 + 3.08 + 3.15} {3 }\\\\S_{avg} = 3.13\rm \ s$

Therefore, the average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

Learn more about Average speed:

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2 years ago

PLEASE HELP ME WITH THIS PROBLEM

valentinak56 [21]

1) The mass of the continent is $2.13\cdot 10^{21} kg$

2) The kinetic energy of the continent is 274.8 J

3) The speed of the jogger must be 2.76 m/s

Explanation:

The continent is a slab of side 5900 km (so the surface is 5900 x 5900, assuming it is a square) and depth 26 km, therefore its volume is:

$V=(36)(4600)^2=7.62\cdot 10^8 km^3 = 7.62\cdot 10^{17} m^3$

The mass of the continent is given by

$m=\rho V$

where:

$\rho = 2790 kg/m^3$ is its density

$V=7.62\cdot 10^{17} m^3$ is its volume

Substituting, we find the mass:

$m=(2790)(7.62\cdot 10^{17})=2.13\cdot 10^{21} kg$

To find the kinetic energy, we need to convert the speed of the continent into m/s first.

The speed is

v = 1.6 cm/year

And we have:

1.6 cm = 0.016 m

$1 year = (365)(24)(60)(60)=3.15\cdot 10^7 s$

So, the speed is

$v=\frac{0.016 m}{3.15 \cdot 10^7 s}=5.08\cdot 10^{-10}m/s$

Now we can find the kinetic energy of the continent, which is given by

$K=\frac{1}{2}mv^2$

where

$m=2.13\cdot 10^{21} kg$ is the mass

$v=5.08\cdot 10^{-10}m/s$ is the speed

Substituting,

$K=\frac{1}{2}(2.13\cdot 10^{21})(5.08\cdot 10^{-10})^2=274.8 J$

The jogger in this part has the same kinetic energy of the continent, so

K = 274.8 J

And its mass is

m = 72 kg

We can write his kinetic energy as

$K=\frac{1}{2}mv^2$

where

v is the speed of the man

And solving the equation for v, we find his speed:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(274.8)}{72}}=2.76 m/s$

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#LearnwithBrainly

3 0

3 years ago

You can carry a 50 N weight up a flight of stairs that is 3 m high in 10 s. Then you carry 40 N up two flights of stairs in 20 s

d1i1m1o1n [39]

You can do 50 and 10 and carry the 30 and it would be in the same power so you’ll have the same energy

5 0

3 years ago

A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer

son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$ rev/sec

$=\frac{2 \pi \times 7.5}{8}$ rad/sec

= 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

$$$=1640 \times (5.89)^2 \times 7.5$

= 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

= 427126.9 x 7.5

= 3,203,451.75 J

6 0

3 years ago