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Vladimir [108]
3 years ago
12

Radium-226 emits alpha (α), beta (β) and gamma (γ) radiation.

Physics
1 answer:
Furkat [3]3 years ago
3 0

Answer:

Explanation:

alpha

Alpha Radiation (α): A large, unstable nucleus decays to produce a smaller, more stable nucleus and an alpha particle (identical to a helium nucleus, ⁴₂He or ⁴₂α).

It has a very high ionizing energy and low penetrating power. It can be stopped by paper skin

Beta Radiation (β): A neutron in an unstable nucleus decays, forming a proton and emitting a beta (β) particle (identical to an electron, ⁰₋₁e or ⁰₋₁b) and resulting in a more stable nucleus.

It has high ionizing energy and penetrating power. It can be stopped by aluminium sheet

Gamma Radiation (γ): An unstable nucleus releases energy in the form of a high energy photon (no mass)to become more stable; this often accompanies other forms of radioactivity.

It has very high penetrating power and very low ionizing energy. It can be stopped by lead block.

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Answer:

t_1 = \frac{v_i}{a_i}

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0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}

Apply the same formula above, with v(t_2) = v_i and a = a_i, and the car is starting from 0 speed,  we have

v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}

As s(t) = vt + \frac{at^2}{2}. After t = t_1 + t_2, the car would have traveled a distance of

s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}

Hence s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}

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Therefore, Δd would be 2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i

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