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3 years ago

Drag and drop each description into the appropriate category:

Physics

2 answers:

Anestetic [448]3 years ago

6 0

Answer:

All of the following are already correctly categorized. This facts about these radioactive decays are true. Further, I have explained below in explanation portion.

Explanation:

Alpha Decay: Yes it cannot penetrate through paper & Yes it emits Alpha particle when an element undergoes the process of Alpha Decay. Here's how:

First of all, we need to understand why a particular element decays? why radioactive decay occurs?

It is because, everything in this universe tends to achieve its stable state. Everything put efforts to be in their stable state. Likewise, when radioactive elements like Uranium or Thorium become unstable they undergo the process of Alpha decay. As a result, they emit energy and alpha particle.

Furthermore, Alpha decay has low penetration power and it has higher mass than Beta Decay and Gamma Decay that's why it cannot travel in long distances. Hence, it cannot pass through a sheet of paper.

Beta Decay: Yes it emits electrons or positrons and Yes it can penetrate through paper but cannot penetrate through a foil. Here's how:

Till now, we are familiar that why the process of decay occurs right? in order to be stable. Like wise Beta Decay occurs and it emits beta particle and energy. The main difference between alpha decay and beta decay is that alpha particle's mass is much greater than beta particle's mass because beta particle is either a negatively charged electron or positron (a positively charged electron or anti-electron) and we know that mass electron is much smaller.

So, Beta Decay has moderate penetrate power and it's mass is much smaller that's why it can travel in longer distances and it can penetrate paper but cannot penetrate into foil.

Gamma Decay: Yes it requires thick lead shield and yes it emits photon. Here's how:

Gamma Decay is a radioactive wave, it only emits energy in form of photons and it has no mass and no charge and but it has very low wavelength but highest energy with great penetration power and that is the reason that it can penetrate paper, it can penetrate foil but it cannot penetrate thick lead shield.

Hope, this will help :)

goldenfox [79]3 years ago

6 0

Answer:

on edge2020 the answers

A. C

B.B

C. C

D.A

E.C

Explanation:

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A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0

3 years ago

A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excit

anygoal [31]

Answer:

a) 23.2 e V

b) energy of the original photon is 36.8 eV

Explanation:

given,

energy at ground level = -13.6 e V

energy at first exited state = - 3.4 e V

A photon of energy ionized from ground state and electron of energy K is released.

h ν₁ - 13.6 = K

K combine with photon in first exited state giving out photon of energy

$h\nu_2 =\dfrac{hc}{\lambda}=\dfrac{12400}{466}$

= 26.6 e V

h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°

K + ( 3.4 ) = 26.6 e V

a) energy of free electron

K = 26.6 - 3.4 = 23.2 e V

b) energy of the original photon

h ν₁ - 13.6 = K

h ν₁ = 23.2 + 13.6

= 36.8 e V

energy of the original photon is 36.8 eV

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Red light is bent the least of all colors as it passes through a prism. What does this tell you about red light? It has a short

Alik [6]

Answer:

Longest wavelength, lowest intensity

Explanation:

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The relatively small, rocky bodies generally found orbiting between mars and jupiter are known as _____. meteoroids satellites c

adelina 88 [10]

The answer that is being described above is the ASTEROIDS. The one that we see floating between Mars and Jupiter is what we call the Asteroid Belt. The asteroid belt comprises of different rocky bodies and they also orbit within the solar system. Hope this helps.

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Taya2010 [7]

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