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Advocard [28]
2 years ago
14

PLEASE HELP ME WITH THIS ONE QUESTION

Physics
1 answer:
Marizza181 [45]2 years ago
7 0

Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>

[A] = Our incognite

k is constant decay:

k = ln 2 / Half-life

k = ln 2 / 4.96 x 10^3 s

k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]

ln[A] = 37.538

[A] = 2.01x10¹⁶ nuclei remain ≈

<h3>c) 2.02 x 10^16 nuclei</h3>
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In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, how many fringes are contained wi
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Answer:

Explanation:

Width of central diffraction peak is given by the following expression

Width of central diffraction peak= 2 λ D/ d₁

where d₁ is width of slit and D is screen distance and λ is wave length.

Width of other fringes become half , that is each of  secondary diffraction fringe is equal to

λ D/ d₁

Width of central interference  peak is given by the following expression

Width of  each of  bright fringe =  λ D/ d₂

where d₂ is width of slit and D is screen distance and λ is wave length.

Now given that the central diffraction peak contains 13 interference fringes

so ( 2 λ D/ d₁)  /  λ  D/ d₂ = 13

then (  λ D/ d₁)  /  λ  D/ d₂ = 13 / 2

= 6.5

no of fringes  contained within each secondary diffraction peak = 6.5

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3 years ago
The amplitude of a mechanical wave shows<br> What?
Anuta_ua [19.1K]
I’m not sure what you are asking
5 0
3 years ago
PHYSICS 50 POINTS PLEASE HELP
tangare [24]

Answer:

One way to look at Newton’s three laws of motion is this:

The third law states what forces are. That is, all forces are interactions between two different objects. If one object is interacting with another, then equal and opposite forces act on each object. So no force acts alone. When you exert a force on something, it is exerting the identical force back on you.

The first and second laws deal with the consequences of the forces that act on an object. The first law says that in the absence of a net force on an object, it simply continues doing whatever it was already doing. If it is at rest, it will remain at rest. If it is in motion, it will continue with that same motion - at constant speed and in the direction it was already traveling.

The second law says what happens if there is a net force on the object. In that case, the object accelerates - either by changing its speed, its direction, or both - in proportion and in the direction of the net force that acts on it. The amount of acceleration depends the object’s mass. That is, the larger the mass the smaller the acceleration for a given net force. The first and second laws can be summarized in the mathematical expression

F = ma

where F is the vector sum of all the forces that act on the object at any given moment (i.e., the net force), m is the mass of the object, and a is the acceleration of the object due to the net force at that moment - and is always in the same direction of the net force.

And notice that in a way, the first law is then “contained” within the second. That is, if the net force is zero on an object, then so is the acceleration. That is, either the object is (still) at rest or, if already in motion, the velocity didn’t change, in either case, the acceleration was zero.

Explanation:

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2 years ago
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Explanation:

Below is an attachment containing the solution.

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3 years ago
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(a)

Electronic configuration is given as follows:

[Kr]4d^{3}

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Thus, electronic configuration of neutral atom is [Kr]4d^{5}5s^{1}.

The atomic number of Kr is 36, thus, total number of electrons become 36+6=42.

This corresponds to element: molybdenum. Thus, the tripositive atom will be Mo^{3+}.

(b) The given electronic configuration is [Kr]5s^{2}4d^{2}.

The atomic number of Kr is 36, thus, total number of electrons become 36+4=40.

This corresponds to element zirconium, represented by symbol Zr.

6 0
3 years ago
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