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iogann1982 [59]
3 years ago
12

Several types of radiation may be emitted during radioactive decay. The equation represents

Physics
2 answers:
vovangra [49]3 years ago
7 0

Answer:

The equation represents the origin of a beta particle from the nucleus of an atom. Beta particles are fast moving electrons; they have a negative charge and originate when a neutron decomposes into a protons, which remains behind in the nucleus, and an electron.

Explanation:

Fittoniya [83]3 years ago
5 0
I think the answer is B
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Suppose you want to make a nested function call (i.e. a call to a function from inside of another function) using a jal rather t
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As we need to use a nested loop in our function,hence push $ra

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8 0
3 years ago
A ferry approaches shore, moving north with a speed of 6.2 m/s relative to the dock. A person on the ferry walks from one side o
jasenka [17]

Speed of Ferry is towards North with magnitude 6.2 m/s

Here if we assume that North direction is along Y axis and East is along X axis then we can say

\vec v_f = 6.2 \hat j

Now a person walk on ferry with speed 1.5 m/s towards east with respect to Ferry

so it is given as

\vec v_{pf} = 1.5 \hat i

also by the concept of relative motion we know that

\vec v_{pf} = \vec v_p - \vec v_f

now plug in all values in it

1.5 \hat i = \vec v_p - 6.2 \hat j

\vec v_p = 1.5 \hat i + 6.2 \hat j

now if we need to find the speed of the person then we need to find its magnitude

so it is given as

v = \sqrt{1.5^2 + 6.2^2}

v = 6.37 m/s

7 0
3 years ago
A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.
Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

\Delta K = \Delta E_{p} = q\Delta V

<u>Where</u>:

K: is the kinetic energy

E_{p}: is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV                                    

\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J

Now, the kinetic energy of the particle as it moves through point B is:

\Delta K = K_{f} - K_{i}

K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!      

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