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3 years ago

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Which states of matter have particles that move independently of one another with very little attraction

1 answer:

Crazy boy [7]3 years ago

6 0

The state of matter that the particles move independently of one another with very little attraction is, I believe, gas

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Mercury is commonly used in thermometer give reasons

evablogger [386]

Answer:

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Explanation:

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which of these rocks are formed by volcanic activity? select the two correct answers. a. limestone b. granite c. marble d. basal

Gennadij [26K]

I think its = b and d

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Radioactive carbon, or C-14, is used to date fossil remains. When C-14 decays, as seen in the nuclear equation, it produces what

olga nikolaevna [1]

The reaction of radiodecay of carbon C-14 is
C-14 --> N-14 + e- + (ve)
where e- is an electron and (ve) is an electron-type antineutrino.
Basically, when the carbon nucleus (atomic number: 6, mass number: 14) decays, a neutron of the nucleus converts into a proton (therefore, the mass number remains the same, 14, but the atomic number increases by 1, therefore it becomes nitrogen) and releases an electron-antineutrino pair.

So, the correct answer is C), N-14.

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Which of the following statements CANNOT be supported by Kepler's laws of planetary motion?

horsena [70]

Answer:

B) A planet's speed as it moves around the sun will not be the same in six months.

Explanation:

A planet's speed as it moves around the sun will not be the same in six months, is a statement that CANNOT be supported by Kepler's laws of planetary motion.

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A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

7 0

3 years ago