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3 years ago

12

Consider three axes of rotation for a pencil: along the lead, at right angles to the lead at the middle, and at right angles to

the lead at one end. Rate the rotational inertias about each axis from smallest to largest.

1 answer:

tensa zangetsu [6.8K]3 years ago

4 0

Answer:

along the lead, at right angles to the lead at the middle, and at right angles to the lead at one end.

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Since all stars begin their lives with the same basic composition, what characteristic most determines how they will differ?

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4. Mass they are formed with

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One complete expression of a waveform beginning at a certain point, progressing through the zero line to the wave’s highest (cre

elixir [45]

Answer:

wavelength.

Explanation:

One complete expression of a waveform beginning at a certain point, progressing through the zero line to the wave’s highest (crest) and lowest (trough) points, and returning to the same value as the starting point is called a is called wavelength. Its can be also defined as the distance between two successive crests or trough points in wave form.

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Which of the following is not a reason that coral reefs are important?

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My Answer: They contain a high diversity of organisms.

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3 years ago

You hold your physics textbook in your hand. (Assume that no other objects are in contact with the book.)

Jet001 [13]

Answer:

1) the correct answer is b and d

2) For force b its direction is vertical up

for the force d its direction is vertical down

3) the correct answers are: a, c and d

4) Force a is vertical down , force c is vertical up and force d is vertical down

5) the correct answer which is b

Explanation:

In this exercise it is asked to identify the forces, fundamentally on the free there are the forces of gravity and the support force of the hand, with these facts we answer the questions

1) the correct answer is b and d

the hand acts on the book with a contact force and the Earth acts on the book with the force of gravity.

2) For force b its direction is vertical up

for the force d its direction is vertical down

3) The forces on the hand are the weight of the book. The force of gravity due to the mass of the hand. As the hand is in balance, there must be a force applied by the arm to keep the hand in position; assuming the hand is in the air, if the hand is resting on the floor the force of the floor on the hand can perform this function

therefore the correct answers are: a, c and d

4) Force a is vertical down

force c is vertical up

force d is vertical down

5) The action and reaction forces are forces of equal magnitude, each applied to one of the bodies, we have

* the force of the hand on the free and its reaction the force of the book on the hand

* The force of the Earth on the book and the hand, giving the weight of each one and the relationship is the force of the book and the hand on the Earth

the correct answer which is b

8 0

3 years ago

A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =

Elis [28]

Answer:

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis is the net horizontal force

$F_v=80062.47\times 10^9$ attractive toward +y axis is the net vertical force

Explanation:

Given:

charge at origin, $Q_0=-3.35\times 10^{-6}\ C$

magnitude of second charge, $Q_2=2.05\times 10^{-6}\ C$

magnitude of third charge, $Q_3=5\times 10^{-6}\ C$

position of second charge, $(x_2,y_2)\equiv(0,4.35)\ cm$

position of third charge, $(x_3,y_3)\equiv(3.1,3.8)\ cm$

<u>Now the distance between the charge at at origin and the second charge:</u>

$d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}$

$d_2=\sqrt{(0-0)^2+(4.35-0)^2}$

$d_2=0.0435\ m$

<u>Now the distance between the charge at at origin and the third charge:</u>

$d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}$

$d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}$

$d_3=0.04904\ m$

<u>Now the force due to second charge:</u>

$F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}$

$F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}$

$F_2=32175.98\times 10^9\ N$ attractive towards +y

<u>Now the force due to third charge:</u>

$F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}$

$F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}$

$F_3=61748.38\times 10^9\ N$ attractive

<u>Now the its horizontal component:</u>

$F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9$

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis

<u>Now the its vertical component:</u>

$F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9$

$F_{3v}=47886.49\times 10^9\ N$ upwards attractive

Now the net vertical force:

$F_v=F_{3v}+F_2$

$F_v=47886.49\times 10^9+32175.98\times 10^9$

$F_v=80062.47\times 10^9$

3 0

3 years ago