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3 years ago

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What characteristics do atoms of carbon-12, carbon-13, and carbon-14 have in common? 1. same number of electrons 2. same number

of protons 3. same number of neutrons and electrons 4. same number of neutrons and protons 5. same number of protons and electrons 6. same mass 7. same number of neutrons

1 answer:

Rasek [7]3 years ago

7 0

Answer:

The most common of these is carbon 12, 13, 14. All of these isotopes have the same atomic number but different mass numbers. Carbon has the atomic number of 6 which means that all isotopes have the same proton number. However, the number of neutrons is different, thus giving different mass numbers.

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If Star A is twice as far as Star B, and they are identical in all other ways, then the

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Explanation:

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You toss a ball in the air. what is the work done by gravity as the ball goes up?

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3 years ago

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea

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Answer:

Q1) Time taking by particle to travel the 40 km wrt. earth = $1.34\times10^{-6}$ sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = $1.285\times10^{-5}$ sec.

Explanation:

Given :

Speed of particle wrt. earth $v=0.99537c$

Distance between where particle is created and earth surface = 40 km

we know that,

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∴ $t = \frac{x}{v}$

$= \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }$

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∴ Thus, time taking by particle to travel the 40 km wrt. earth $t = 1.34\times10^{-6} sec$

According to the lorentz transformation,

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Where $l =$ improper length, $l_{o} =$ proper length (distance measured wrt. rest frame) = 40 km

$l = 40 \sqrt{1-\frac{v^2}{c^2 }$

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$\Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }$

Where $\Delta t =$ improper time (wrt. earth frame time) $=1.34\times10^{-6} sec$ , $\Delta t _{o} =$ proper time (wrt. particle frame).

$1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}$

$\Delta t_{o} = 1.285 \times10^{-5} sec$

Thus, the time taking by particle to travel from where it is created to the surface of the earth $= 1.285 \times10^{-5} sec$ .

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3 years ago

Read 2 more answers

George walks to a friend's house. He walks 850 meters north before he realizes he walked too far.

anyanavicka [17]

Answer:

<h2>44 m/s</h2>

Explanation:

In this problem we are expected to calculate the velocity of Georges movements.

Given data

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v= 1100/ 25= 44 m/s

Hence the velocity in m/s is 44

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3 years ago