All categories

3 years ago

13. A nuclear decay chain represents

Physics

1 answer:

murzikaleks [220]3 years ago

5 0

Answer:

a series of decays producing sequentially more stable nuclei

Explanation:

A nuclear decay chain represents a series of decays producing sequentially more stable nuclei.

For every atomic nucleus, there is a specific neutron/proton ratio which ensures the stability of the nucleus.
Any nucleus with a neutron/proton combination different from its stability ratio (i.e either too many neutrons or too many protons) will be unstable and split into one or more other nuclei with the attendant emission of small particles of matter.
The decay process is a continuous chain reaction steps.
Until stability of the nucleus is attain, it continues.

You might be interested in

I need an answer for this plzz!!<br>number 2 <br>anybody can help ??

Anuta_ua [19.1K]

2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
$a \: = \: \frac{dv}{dt} \: = \: \frac{25 \: \frac{m}{s} }{50 \: s} \: = \: 0.5 \: \frac{m}{ {s}^{2} }$
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
$2ax \: = \: {v}^{2} \: - \: {u}^{2}$
$x \: = \: \frac{ {v}^{2} \: - \: {u}^{2} }{2a} \: = \: \frac{{(25 \: \frac{m}{s})}^{2} \: - \: {(0 \: \frac{m}{s} )}^{2} }{2 \: \times \: 0.5 \: \frac{m}{ {s}^{2} } } \: = \: 625 \: m$
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.

8 0

3 years ago

A tea kettle is boiling on the stove, and all of a sudden the whistle in the spout of the kettle starts shrieking. Steam rising

fenix001 [56]

Answer:

The steam represents evaporation.

Explanation:

4 0

2 years ago

Read 2 more answers

calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

3 years ago

Static cling is an example of

cestrela7 [59]

I would say c hope it helps:)

3 0

3 years ago

A mystery element has three isotopes of the following masses and percent abundances: 93.597 amu at 23.63 % abundance, 96.191 amu

iogann1982 [59]

Answer:

92.397amu

Explanation: The exact amu of the mystery element is obtained by multiplying the relative abundance of each individual isotope by its respective amu and then summing the results.

The sum of the total relative abundance for all the isotopes should be 100%.

However, the relative abundance of the isotope with 95.502amu is not given; therefore to obtain it we subtract the sum of the known relative abundances from 100% as follows:

Relative abundance of isotope with 95.502amu = 100-(23.63+30.53) = 42.84%

8 0

3 years ago