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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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A straight wire of length 0.56 m carries a conventional current of 0.4 amperes. What is the magnitude of the magnetic field made
djverab [1.8K]

Answer:

The magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.

Explanation:

Given;

length of the straight wire, L = 0.56 m

conventional current, I = 0.4 A

distance of magnetic field from the wire, r = 2.6 cm = 0.026 m

To determine magnitude of magnetic field made by current in the wire, we will apply Bio-Savart Law;

B = \frac{\mu_o}{4\pi r} \frac{LI}{\sqrt{r^2 +(L/2)^2} } \\\\B = \frac{4\pi *10^{-7}}{4\pi *0.026} \frac{0.56*0.4}{\sqrt{(0.026)^2 +(0.56/2)^2} }\\\\B = 3.846*10^{-6}(0.7966)\\\\B = 3.064*10^{-6} \ T

Therefore, the magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.

3 0
3 years ago
A thin, flat washer is a disk with an outer diameter of 14 cm and a hole in the center with a diameter of 7 cm. The washer has a
Kryger [21]

Answer:

2583.9 N/C

Explanation:

Parameters given:

Outer diameter = 14 cm

Outer radius, R = 7cm = 0.07m

Inner diameter = 7 cm

Inner radius, r = 3.5 cm = 0.035m

Charge of washer = 8 nC = 8 * 10^(-9)C

Distance from washer, z = 33 cm = 0.33m

The electric field due to a washer (hollow disk) is given as:

E = k * σ * 2π [ 1 - z/(√(z² + R²)]

Where σ = charge per unit area

σ = q/π(R² - r²)

σ = 8 * 10^(-9) /(π*(0.07 - 0.035)²)

σ = 2.077 * 10^(-6) C/m²

=> E = 9 * 10^9 * 2.077 * 10^(-6) * 2π * [1 - 0.33/(√(0.33² + 0.07²)]

E = 117.467 * 10^3 * (1 - 0.978)

E = 117.467 * 10^3 * 0.022

E = 2583.9 N/C

6 0
3 years ago
A ball is ejected to the right with an unknown horizontal velocity from the top of a pillar that is 50 meters in height. At the
dimulka [17.4K]

Answer:

15.67 m/s

Explanation:

The ball has a projectile motion, with a horizontal uniform motion with constant speed and a vertical accelerated motion with constant acceleration g=9.8 m/s^2 downward.

Let's consider the vertical motion only first: the vertical distance covered by the ball, which is S=50 m, is given by

S=\frac{1}{2}gt^2

where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(50 m)}{9.8 m/s^2}}=3.19 s

Now we now that the ball must cover a distance of 50 meters horizontally during this time, in order to fall inside the carriage; therefore, the velocity of the carriage should be:

v=\frac{d}{t}=\frac{50 m}{3.19 s}=15.67 m/s

8 0
3 years ago
What is common between the quantities of area, density and speed?​
Rainbow [258]

Answer:

They are all s

calar quantities

5 0
2 years ago
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if a sample that is 4860 years old has 50 radium atoms remaining, how many atoms were in the original sample?
snow_lady [41]
243000, take the 4,860 and times it by 50
4 0
3 years ago
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