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gavmur [86]
3 years ago
9

After a nucleus undergoes radioactive decay, its new mass number is:

Physics
2 answers:
Ivanshal [37]3 years ago
7 0
Radioactive "decay" means particles and stuff shoot OUT of a nucleus.
After that happens, there's less stuff in the nucleus than there was before.
So the new mass number is always less than the original mass number.
lapo4ka [179]3 years ago
7 0

<u>Answer:</u> The new mass number is never more than its original mass number

<u>Explanation:</u>

Radioactive decay is defined as the process in which an unstable nuclei breaks down into stable nuclei via various methods.

An isotope undergoes a radioactive decay to attain stability.

There are many decay processes by which a parent nucleus can undergo decay. They are:

Alpha decay is defined as the decay process in which alpha particle is released. In this process, a heavier nuclei decays into a lighter nuclei. The alpha particle released carries a charge of +2 units and a mass of 4 units.

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha

Beta decay is defined as the decay process in which a neutron gets converted to a proton and an electron.  In this decay process, beta particle is emitted. The emitted particle carries a charge of -1 units and has a mass of 0 units. The released beta particle is also known as electron.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

Gamma decay is defined as the decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. This decay releases \gamma -radiations. This process does not change the mass number.

_A^Z\textrm{X}^*\rightarrow _A^Z\textrm{X}+_0^0\gamma

For every decay process, the mass number will either remain same or the mass number of new isotope will be less than the parent isotope but not more than the original mass number.

Hence, the new mass number is never more than its original mass number

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A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a
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Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N

Each thruster has to applied a force of 294.5N in tangential direction

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3 years ago
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ikadub [295]

Answer:

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Explanation:

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