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svp [43]
2 years ago
8

The nucleus of the polonium isotope 214 Po (mass 214 u) is radioactive and decays by emitting an alpha particle (a helium nucleu

s with mass 4 u). Laboratory experiments measure the speed of the alpha particle to be 1.97×107 m/s.
Assuming the polonium nucleus was initially at rest, what is the recoil speed of the nucleus that remains after the decay?
Physics
2 answers:
amid [387]2 years ago
6 0

Answer:

368224.29906 m/s

Explanation:

M = Mass of Polonium nucleus = 214 u

V = Velocity of nucleus

m = Mass of Helium nucleus = 4 u

v = Velocity of alpha particle = 1.97\times 10^7\ m/s

In this system the momentum is conserved

MV+mv=0\\\Rightarrow MV=-mv\\\Rightarrow 214V=-4\times 1.97\times 10^7\\\Rightarrow V=\dfrac{-4\times 1.97\times 10^7}{214}\\\Rightarrow V=368224.29906\ m/s

The recoil speed of the nucleus that remains after the decay is 368224.29906 m/s

Maksim231197 [3]2 years ago
4 0

Answer:

368224.29906

Explanation:

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3 0
3 years ago
Which best describes the current atomic theory?
ivolga24 [154]
Choice-C is a correct statement.
6 0
2 years ago
Read 2 more answers
What was the long-term effect of the Northwest Ordinance of 1787?
victus00 [196]

Answer:

Established a government for the Northwest Territory, outlined the process for admitting a new state to the Union, and guaranteed that newly created states would be equal to the original thirteen states

Explanation:

Goooogle, so I hope this helps somewhat

Also, isn't this a History question? You put physics lol

4 0
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A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p
alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

F=qvB

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

qvB = \frac{mv^2}{r}

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v=\frac{qB}{mr}

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

\frac{2\pi r}{T}=\frac{qB}{mr}

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T=\frac{2\pi m r^2}{qB}

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) a. f/10

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f=\frac{1}{T}

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f is the frequency

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We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

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6 0
3 years ago
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ANTONII [103]
A water wave is an example of a mechanical wave. A wave that can travel only through matter is called a mechanical wave.
6 0
2 years ago
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