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Alika [10]
3 years ago
9

In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.

Physics
1 answer:
vaieri [72.5K]3 years ago
4 0

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X

^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

^{211}_{83}Bi\Rightarrow ^{211}_{83+1}X

^{211}_{83}Bi\Rightarrow ^{211}_{84}Po

The resulting nucleus is Po.

(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

^{22}_{11}Na\Rightarrow ^{22}_{11-1}X

^{22}_{11}Na\Rightarrow ^{22}_{10}Ne

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

^{98}_{43}Tc\Rightarrow ^{98}_{43}Tc

The resulting nucleus is Tc.

Hence, (A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

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Artist 52 [7]

Answer:

The height of the water column = 1.62405\overline{30} × 10⁻¹ m

Explanation:

The air cavity in the Coke bottle = 0.220 m deep

The fundamental (frequency) it plays when water is added to shorten the column and it is blown across the top, f = 528 Hz

The given speed of sound in air, v = 343 m/s

We note that the air cavity in the coke bottle is equivalent to a tube closed at one end

The fundamental frequency for a tube closed at one end, 'f', is given as follows;

f = v/(4·L) = v/λ

Where;

L = The height of the water column

λ = The wavelength of the wave

∴ 4·L = v/f = (343 m/s)/(528 Hz) = 0.6496\overline{21} m

∴ L = 0.6496\overline{21} m/4 = 0.162405\overline{30} m

The height of the water column = 1.62405\overline{30} × 10⁻¹ m.

4 0
3 years ago
Read 2 more answers
A student wishes to work out how much power she uses to lift her body when climbing a
Bess [88]

Answer

i'm not 100% sure but 1764

Explanation:

Work done = gravitational potential energy

Gravitational potential energy = mass(kg) × height(m) × gravitational field strength(N/kg)

We can assume that the student is on earth so the gravitational field strength is 9.8N/kg

So work done = 60 × 3×9.8

=1764

(if you need help calculating power but if you do just divide your answer by 12 and you will get 147)

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2 years ago
Children are told to avoid standing too close to a rapidly moving train because they might get sucked under it. Is this possible
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Answer:

no its not like the undertow in the ocean

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4 0
3 years ago
How many protons would the element with the atomic number 10 contain?
galben [10]
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3 0
3 years ago
A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga
tino4ka555 [31]

Answer

given,

diameter,d₁ = 7.5 cm

               d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₂² v₂

 v_1= \dfrac{r_2^2}{r_1^2} v_2

 v_1= \dfrac{2.25^2}{3.75^2} v_2

 v_1= 0.36 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)

 32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)

 v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}

 v_2=\sqrt{16.084}

       v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)²  x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0
3 years ago
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