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kati45 [8]
3 years ago
11

What is the science behind the making of pop-up books?

Physics
1 answer:
IRINA_888 [86]3 years ago
6 0

Answer

The moveable parts of the pop up book are of often cut out by hand and are folded and glued by hand upon the printed pages. The cover is glued or sewn to the lining. Front and backs are often made up from board, which is just a heavier gauge paper than is used for the pages.

Explanation

Hope that this helps you and have a great day:)

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How do sea surface temperatures affect evaporation rate?
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<span>Answer: The temperature doesn't affect the evaporation rate, but affects on how much of water a parcel of air can contain when saturated which is known by the absolute humidity. Hurricanes are usually happening when the temperature of the sea water west of the Cape Verde islands is over 27 degrees Celsius. If ahead of the path of a hurricane, the sea water temperature drops then it will be less moisture in the air and perhaps the hurricane will fade out. But it is not as simple. How strong a tropical storm is is relative to the difference of temperture between ground level and the top of the troposphere. The greater the difference, the faster the air will rise and the deeper the pressure will be, forcing surrounding air to rush in, thus forming a hurricane force wind. Then there is the fact that the wet adiabatic lapse rate is about half that of dry air. It means that rising moist air cools down slower and therefore rises higher. Hence water is the true fuel of bad weather. But it can't be isolated from the fact that the difference of temperature must be great too. What we often forget is that the tropopause (the border to the stratosphere) is much higher over the equator and therefore, much colder than e.g. the poles.</span>
8 0
3 years ago
Just the answer (PLSS)
Salsk061 [2.6K]

The miracle year for Albert Einstein  was the year 1905 within which he published so many renowned papers.

<h3>When was Einstein miracle year?</h3>

The miracle year for Albert Einstein  was the year 1905 within which he published so many renowned papers in a short time and became very popular.

His mindset in that year was one that challenged the orthodox explanations and sought to think outside the box.

Learn more about Albert Einstein:brainly.com/question/2964376

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8 0
2 years ago
Indicate on the chart whether you would classify each model as representing an element, compound, or mixture.
Ne4ueva [31]

Answer:

1 compound

2 mixture

3 elements

4 0
3 years ago
Read 2 more answers
How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'
Serhud [2]

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity (K.E.=\frac{1}{2}mv^2). If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height(P.E.=mgh). If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.


4 0
3 years ago
Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm
Ghella [55]

Answer:

Total electric potential, V=1.32\times 10^6\ volts

Explanation:

It is given that,

First charge, q_1=3\ \mu C=3\times 10^{-6}\ C

Second charge, q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C

Distance of first charge from origin, r_1=1.15\ cm=0.0115\ m

Distance of second charge from origin, r_2=2\ cm=0.02\ m

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}

V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})

V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})

V = 1321826.08 V

or

V=1.32\times 10^6\ volts

So, the total electric potential at the origin is 1.32\times 10^6\ volts. Hence, this is the required solution.

3 0
2 years ago
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