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Alex777 [14]
3 years ago
9

When the daughter nucleus produced in a radioactive decay is itself unstable, it will eventually decay and form its own daughter

nucleus. If the newly formed daughter nucleus is also unstable, another decay will occur, and the process will continue until a nonradioactive nucleus is formed. Such a series of radioactive decays is called a decay chain.
A good example of a decay chain is provided by 232 90Th, a naturally occurring isotope of thorium.

What is the energy Q released in the first step of the thorium-232 decay chain? The atomic mass of 232 90Th is 232.038054 u and the atomic mass of 228 88Ra is 228.0301069 u.

Answer in (MeV) and show your work
Physics
1 answer:
Dmitrij [34]3 years ago
3 0

Answer:

4.981 MeV

Explanation:

The quantity of energy Q can be calculated using the formula

Q = (mass before - mass after) × c²

Atomic Mass of thorium = 232.038054 u, atomic of Radium = 228.0301069 u and mass of Helium = 4.00260. The difference of atomic number and atomic mass  between the thorium and radium ( 232 - 228)  and ( 90 - 88)  show α particle was emitted.

1 u = 931.494 Mev/c²

Q = (mass before - mass after) × c²

Q = ( mass of thorium - ( mass of Radium + mass of Helium ) )× c²

Q = 232.038054 u - ( 228.0301069 + 4.00260) × c²

Q = 0.0053471 u × c²

replace 1 u = 931.494 MeV/ c²

Q = 0.0053471 × c² × (931.494 MeV / c²)

cancel c²  from the equation

Q = 0.0053471 × 931.494 MeV = 4.981 MeV

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The energy flow per unit time per unit area (S) of an electromagnetic wave has an average value of 310 mW/m2. The maximum value
Aleksandr-060686 [28]

Answer:

5.096*10^-8

Explanation:

Given that

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Emax = √[(2 * 0.310 * 3*10^8 * 4π*10^-7)]

Emax = √233.7648

Emax = 15.289

Now, with our value of maximum electromagnetic wave gotten, we divide it by speed of light to get our final answer

15.289 / (3*10^8) = 5.096*10^-8 T

Suffice to say, The maximum value of the magnetic field in the wave is closest to 5.096*10^-8

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2 years ago
You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-hi
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Answer:

m = maximum mass of the coaster = 410 kg

d = maximum spring compression = 2.3 m

h = maximum height of the track = 11 m

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g = acceleration by gravity = 9.8 m/s²

k = spring constant (without safety margin) = ?

K = spring constant (with safety margin) = ?

V = maximum speed of the coaster = ?

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PEg = 44198 J

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k = 16710 N/m

Adding 14% to that value, you get:

K = 1.14 (16710 N/m)

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When fully compressed, the elastic potential energy stored in the spring is:

PEe = K d² / 2

PEe = (19045 N/m) (2.3m)² / 2

PEe = 51326 J

The difference in height between the starting point and the lowest point of the track is:

Δh = H - h

Δh = (19 m) - (11 m)

Δh = 8 m

So the initial gravitational potential energy of 330 kg coaster, relative to the lowest point, is

PEg = m g Δh

PEg = (340 kg) (9.8 m/s) (8 m)

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The total energy of the coaster at its starting point (again, relative to the lowest point) is:

TE = PEe + PEg

TE = (51326J) + (26656 J)

TE = 77982J

At the lowest point of the track, all that energy is converted to kinetic energy, so the speed at that point will be:

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The upward force exerted by the elevator floor on the passenger is mass times acceleration, 7.6 times mass of passenger.

To learn more about force refer to the link:

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