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maxonik [38]
3 years ago
12

Which scientific design has the fewest limitations?

Physics
1 answer:
AleksAgata [21]3 years ago
4 0
USING DISSECTION TO STUDY THE GENERAL STRUCTURE OF LILY FLOWERS.

Other choices are very hard to do. Maintaining a salt-water aquarium is very hard while going to an arctic tundra or space is impractical. 
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A stone is thrown upward at an angle. what happens to the horizontal component of its velocity as it rises? as it falls?
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What is a example of a real life scientific inquiry problem
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What is an example of how you can use scientific inquiry to solve a real life problem.
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3 years ago
A battery is used to charge a parallel-plate capacitor, after which it is disconnected. Then the plates are pulled apart to twic
Igoryamba

Answer: C.

Explanation:

For a parallel-plate capacitor where the distance between the plates is d.

The capacitance is:

C = e*A/d

You can see that the distance is in the denominator, then if we double the distance, the capacitance halves.

Now, the stored energy can be written as:

E = (1/2)*Q^2/C

Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:

E = (1/2)*Q^2*d/(e*A)

e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.

Then we can define:

K = (1/2)*Q^2/(e*A)

And now we can write the energy as:

E = K*d

Then the energy is proportional to the distance between the plates, this means that if we double the distance, we also double the energy.

8 0
3 years ago
Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po
zvonat [6]

Answer:

0.293I_0

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

I_1=\frac{I_0}{2}

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

I_2=I_1 cos^2 \theta

where

\theta is the angle between the axes of the two polarizer

Here we have

\theta=40^{\circ}

So the intensity after the 2nd polarizer is

I_2=I_1 (cos 40^{\circ})^2=0.587I_1

And substituting the expression for I1, we find:

I_2=0.587 (\frac{I_0}{2})=0.293I_0

5 0
3 years ago
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