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Mrrafil [7]
3 years ago
7

During nuclear decay, a new isotope is created. How is the nucleus of the new isotope different from the parent if the parent is

otope has undergone alpha decay?
Physics
2 answers:
Veseljchak [2.6K]3 years ago
6 0

Answer:

the difference in parent nuclei and daughter nuclei is that daughter nuclei is having smaller mass number and by 4 and smaller atomic number by 2

_z^AX ----> _{z-2}^{A-4}Y + _2^4He

Explanation:

Alpha nucleus is the helium nuclei which is having 2 protons and 2 neutrons

So we will have

\alpha = _2^4He

now we know that in all radioactive decay mass number and atomic number will remain constant on product and reactant side

So we have

X ------> Y + \alpha

now by mass number conservation and atomic number conservation we have

_z^AX ----> _{z-2}^{A-4}Y + _2^4He

so the difference in parent nuclei and daughter nuclei is that daughter nuclei is having smaller mass number and by 4 and smaller atomic number by 2

STALIN [3.7K]3 years ago
3 0

Answer:

Alpha decay will produce a daughter nucleus with more protons and beta decay will produce a daughter nucleus with fewer protons than the parent nucleus has.

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A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object
Elina [12.6K]

Answer:

2f

Explanation:

The formula for the object - image relationship of thin lens is given as;

1/s + 1/s' = 1/f

Where;

s is object distance from lens

s' is the image distance from the lens

f is the focal length of the lens

Total distance of the object and image from the lens is given as;

d = s + s'

We earlier said that; 1/s + 1/s' = 1/f

Making s' the subject, we have;

s' = sf/(s - f)

Since d = s + s'

Thus;

d = s + (sf/(s - f))

Expanding this, we have;

d = s²/(s - f)

The derivative of this with respect to d gives;

d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²

Equating to zero, we have;

(2s/(s - f)) - s²/(s - f)² = 0

(2s/(s - f)) = s²/(s - f)²

Thus;

2s = s²/(s - f)

s² = 2s(s - f)

s² = 2s² - 2sf

2s² - s² = 2sf

s² = 2sf

s = 2f

8 0
3 years ago
Choose the word that BEST completes the following sentence. Saturn’s rings have been assigned a letter, based on their location.
puteri [66]

Answer:

The correct answer would be Saturn's Cassini Division.

Explanation:

Read about it here.

https://caps.gsfc.nasa.gov/simpson/kingswood/rings/

Hope this helps! :)

5 0
3 years ago
U need help so can some one help me
GrogVix [38]

Answer:

are sure this is a question

7 0
3 years ago
A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
disa [49]
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, \mu m g. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
ma=-\mu m g
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
2aS=v_f^2-v_i^2
where v_f=0 is the final speed of the skier and v_i=4.75 m/s is the initial speed. Substituting numbers, we find:
S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m
5 0
3 years ago
A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55
Anit [1.1K]

Explanation:

We have,

Mass of a baseball is 0.147 kg

Initial velocity of the baseball is 44.5 m/s

The ball is moved in the opposite direction with a velocity of 55.5 m/s

It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Change in momentum,

\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s

Impulse = 14.7 kg-m/s

Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s

4 0
3 years ago
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