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lubasha [3.4K]
3 years ago
8

In beta decay, what is emitted?

Physics
2 answers:
FrozenT [24]3 years ago
8 0
An electron is emitted in both positive and negative beta decay, although the positive one is called positron emission.
bagirrra123 [75]3 years ago
4 0

Answer:The correct answer is an electron.

Explanation:

Beta decay : In this process, a neutron is converted into a proton and an electron. In this decay, the atomic number is increased by 1 unit.

General representation of beta decay :

_Z^A\textrm{X}\rightarrow _{Z+1}^{A}Y+_{-1}^0e

Where as:

Alpha decay : When a larger nuclei decays into smaller nuclei by releasing alpha particle. In this process, the mass number and atomic number is reduced by 4 and 2 units respectively.

General representation of alpha decay :

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}Y+_2^4\alpha

Gamma decay : In this process, the unstable nuclei transformed into stable nuclei by releasing gamma rays. In this process, the mass number remains same.

General representation of gamma decay :

_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma.

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(a) The ball has a final velocity vector

\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j

with horizontal and vertical components, respectively,

v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}

v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}

The horizontal component of the ball's velocity is constant throughout its trajectory, so v_{x,i}=v_{x,f}, and the horizontal distance <em>x</em> that it covers after time <em>t</em> is

x=v_{x,i}t=v_{x,f}t

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s

The vertical component of the ball's velocity at time <em>t</em> is

v_{y,f}=v_{y,i}-gt

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}

So, the initial velocity vector is

\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j

which carries an initial speed of

\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}

and direction <em>θ</em> such that

\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

y=v_{y,i}t-\dfrac12gt^2

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}

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A 150kg motorcycle starts from rest and accelerates at a constant rate along with a distance of 350m. The applied force is 250N
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Note that;

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Equation (i) then becomes;

F = F_{A} - μmg        -------------------(ii)

<em>Given from question;</em>

m = mass of motorcycle = 150kg

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Substitute these values into equation (ii) as follows;

F = 250 - (0.03 x 150 x 10)

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