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ololo11 [35]
3 years ago
7

Trial 1: Get a textbook and put a sheet of paper on top of it. Fold the paper as needed to keep the paper from sticking over the

edge of the book.Hold the textbook with the paper on top, horizontally about waist high.Drop the book and paper so that they hit the floor flat. Record your observations.Trial 2: With the book in one hand and the paper in the other, drop the book and paper simultaneously from the same height. Record your observations.
Physics
1 answer:
siniylev [52]3 years ago
3 0

Answer:

1)  the two objects reach the floor at the same time.

2)the book reaches the floor much earlier than the foil

In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more

Explanation:

This interesting experiment has the following results

1) first case. Sheet on top of book

In this case the two objects reach the floor at the same time.

This shows that the acceleration in the two objects is the same and we call it the acceleration of gravity.

The speed of the body increases as it goes down linearly.

This occurs because the book that receives air resistance is much heavier, so the resistance has almost no effect on its movement, the sheet does not have the air resistance because it goes down next to the book.

2) second case. Book and sheet next to each other.

In this case the book reaches the floor much earlier than the foil.

This is because the resisting force of the air has almost no effect on the book and its movement is little affected by this force.

In the case of the blade, it has very little weight, therefore as its speed increases, the resistance force of the air rapidly equals the weight of the blade.

           W_sheet - fr = 0

so after this, since the acceleration is zero, it goes down at constant speed, this speed is called the terminal velocity.

In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more.

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In an experiment, a large number of electrons are fired at a sample of neutral hydrogen atoms and observations are made of how t
kakasveta [241]

Answer:

N = 1036 times

Explanation:

The radial probability density of the hydrogen ground state is given by:

p(r) = \frac{4r^{2} }{a_{0} ^{3} } e^{\frac{-2r}{a_{0} } }

p(\frac{a_{0} }{2} ) = \frac{4(\frac{a_{0} }{2} )^{2} }{a_{0} ^{3} } e^{\frac{-2(\frac{a_{0} }{2} )}{a_{0} } }

p(2a_{0} ) = \frac{4(2a_{0}) ^{2} }{a_{0} ^{3} } e^{\frac{-4a_{0} }{a_{0} } }

N = 1300\frac{p(2a_{0}) }{p(\frac{a_{0} }{2} )}

N = 1300\frac{(2a_{0}) ^{2}e^{\frac{-4a_{0} }{a_{0} } }  }{(\frac{a_{0} }{2} )^{2} e^{\frac{-a_{0} }{a_{0} } }}

N = 1300(16) e^{-3}

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6 0
3 years ago
What is the distance to a star whose parallex is 0.1 sec?
Arte-miy333 [17]

Answer:

30.86\times 10^{13} km

Explanation:

Given the parallex of the star is 0.1 sec.

The distance is inversely related with the parallex of the star. Mathematically,

d=\frac{1}{P}

Here, d is the distance to a star which is measured in parsecs, and P is the parallex which is measured in arc seconds.

Now,

d=\frac{1}{0.1}\\d=10 parsec

And also know that,

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2 years ago
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A solid-propellant rocket has chamber pressure of 6.35 atm with propellant density of 3.8 g/cm3 and burn area of 975 cm2 . Find
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Answer:

Explanation:

Given:

P = 6.35 atm

= 1.01 × 10^5 × 6.35

= 6.434 × 10^5 N/m^2

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M = 320 kg

Since the propellant volume is equal to the cross sectional area, As times the fuel length, the volumetric propellant consumption rate is the cross section area times the linear burn rate, bs , and the instantaneous mass flow rate of combustion, ms gases generated is equal to the volumetric rate times the fuel density, D

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t = sqrt(3.705 × 10^6/6.43 × 10^5)

= 2.4 s

4 0
3 years ago
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