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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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g If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resi
Anettt [7]
<h2>Question:</h2>

In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?

Answer:

9.1Ω

Explanation:

The circuit diagram has been attached to this response.

(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e

\frac{1}{R_X} = \frac{1}{R_1} + \frac{1}{R_2}

=> R_{X} = \frac{R_1 * R_2}{R_1 + R_2}             ------------(i)

From the question;

R1 = 3Ω,

R2 = 7Ω

Substitute these values into equation (i) as follows;

R_{X} = \frac{3 * 7}{3 + 7}

R_{X} = \frac{21}{10}

R_{X} = 2.1Ω

(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.

Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e

R = Rₓ + R3

Rₓ = 2.1Ω

R3 = 7Ω

=> R = 2.1Ω + 7Ω = 9.1Ω

Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω

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A) amperes

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Read 2 more answers
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