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3 years ago

Which change occurs when an atom undergoes decay?

1 answer:

6 0

<span>When an atom undergoes decay its nucleus becomes unstable because it does not have a balance of neurons and protons. This means the atoms try to find a confirmation in which it is stable. In doing so, it is possible that the atom changes its element, and it frees a huge amount of energy in order to balance itself. This is the base of radioactivity.</span>

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Select the correct answer.

chubhunter [2.5K]

I believe it’s self-referent encoding

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2 years ago

What do we call the distance labeled from A to B and what could we do to the note played on an instrument to change that distanc

Korvikt [17]

Correct answer choice is :

A) From A to B is known as the wavelength and changing the pitch of the note will change its length.

Explanation:

The amount or quantity of period within two things, points, lines, etc. the state or fact of existing separate in space, as of one thing from another; remoteness. a linear amount of space: Seven miles is a distance too great to walk in an hour. Distance is a scalar quantity describing the interval in two points. It is just the measure of the interval.

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3 years ago

Read 2 more answers

Scott travels north 5 miles, then goes west 3 miles, and then goes south for 2 miles.

Yuki888 [10]

Scott traveled 10 miles

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2 years ago

Find the energy u of the capacitor in terms of c and q by using the definition of capacitance and the formula for the energy in

gtnhenbr [62]

The formula for the energy in a capacitor , u in terms of q and c is q²/2c

<h3>What is the energy of a capacitor?</h3>

The energy of a capacitor u = 1/2qv where

q = charge on capacitor and
v = voltage across capacitor.

<h3>What is the capacitance of a capacitor?</h3>

Also, the capacitance of a capacitor c = q/v where

q = charge on capacitor and
v = voltage across capacitor.

So, v = q/c

<h3>The formula for energy of the capacitor in terms of q and c</h3>

Substituting v into u, we have

u = 1/2qv

= 1/2q(q/c)

= q²/2c

So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c

Learn more about energy in a capacitor here:

brainly.com/question/10705986

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2 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$