An experiment is designed to compare the differences in learning outcomes

Physics

1 answer:

MissTica2 years ago

6 0

Answer:

B. Randomly assigning students to the two different groups

Explanation:

The bias can be controlled by making use of the randomized block design where we divide the subject into subgroups also called blocks and randomly assign treatment conditions to each block.

In this case, the blocks or sub groups are learning math from a video game and learning it from traditional classroom activities.

So the best way to control the bias is by randomly assigning students to the different groups.

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If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

3 years ago

A bicycle pump contains 200 cm3 of air and is connected to a bicycle tyre. The volume of the tyre is 800 cm3. The pressure of th

Aleksandr [31]

Answer:

The total volume of the air is 1000 cubic centimeters.

Explanation:

Since the bicycle pump and the bicycle tyre have the same pressure, then the total volume of the air is the sum of the volume of each element, then we translate this into the following artihmetical expression:

$V = 200\,cm^{3}+800\,cm^{3}$