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3 years ago

What is formed from nuclear decay?

Physics

1 answer:

Digiron [165]3 years ago

3 0

Answer:

Energy and another nucleus of the new element/particle.

Explanation:

A nuclear decay is an radioactive decay where the nucleus of an atom becomes unstable and loses energy through emitting radiations.For example, there is an alpha decay that release an alpha particle called helium-4, a beta decay that involve releasing a beta particle which normally occurs in a nucleus that has more neutrons than protons.Other decays are the gamma decay,electron capture and the spontaneous fission.Nuclear decay produces a lot of energy and radiations that could harm human cells if exposed.

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Crest : trough :: compression : _____ A. frequency B. amplitude C. rarefaction D. wavelength

9966 [12]

Rarefraction.

Crest- tallest spot on transverse wave.

Trough- shortest point on transverse wave.

Compression - spot on a compressional wave where the waves are closer together.

Rarefraction - spot on a compressional wave where the waves are farther apart.

3 0

3 years ago

A conducting rod is moving through a magnetic field, as in the drawing. The magnetic field strength is 0.65 T, and the speed of

Contact [7]

Explanation:

The given data is as follows.

Magnetic field strength (B) = 0.65 T

Speed (v) = 2.3 m/s

Induced emf (E) = ?

Formula for emf induced at the ends of the rod of length L which is moving with a speed of v is as follows.

E = BvL

Putting the given values into the formula as follows.

$E_{1} = BvL$

= $0.65 T \times 2.3 m/s \times L$

= 1.495 L .............. (1)

When magnetic field is changed to $B_{2}$ = 0.48 T

Now, we assume that the speed be $v_{2}$ to get the emf $E_{2} = E_{1}$ .

Then, $0.48 T \times v_{2} \times L = 0.65 T \times 2.3 m/s \times L$

$v_{2}$ = 3.11 m/s

Therefore, we can conclude that the speed v of the rod be adjusted to reestablish the emf induced between the ends of the rod at its initial value is 3.11 m/s.

6 0

3 years ago

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

$\omega=\frac{2\pi}{T}$

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$

We can use the chain rule again to find this derivative so we get:

$y" =-\frac{5\pi}{4}sin(\frac{\pi}{4} t)*(\frac{pi}{4})$

so when simplified we get:

$y"=-\frac{5\pi^{2}}{16}sin(\frac{\pi}{4} t)$

its amplitude is:

$A=\frac{5\pi^{2}}{16}$

so its maximum acceleration is:

$a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

7 0

3 years ago

The strength of a magnetic field around a wire carrying a current of 20 A is 0.004 T. What is the strength of the

Alex777 [14]

Answer:

D. 0.008 T

Explanation:

Eng 2021

7 0

3 years ago

Question 1 (1 point)

IrinaK [193]

momentum
Yes, if the elephant is standing still.
Fullback
impulse acting on it.
2.25 N∙s
A cannon firing.
Inelastic
it stays the same
When the cue ball contacts the other balls, momentum is transferred causing them to gain momentum and speed.
less than 3 m/s

<h3><u><em>these are all correct i got an 100%</em></u><em><u> </u></em></h3>

8 0

3 years ago