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PtichkaEL [24]
2 years ago
8

Calculate the radiation dosage (in grays) for a 63‑kg person who is exposed for 5.0 s to a 3.0 Ci source of beta radiation. Assu

me that 100% of the radiation is absorbed and that each beta particle has an energy of 1.5×10−13 J.
Physics
1 answer:
Ierofanga [76]2 years ago
5 0

Answer:

1.32142x10^-3 Gy

Explanation:

Mass m = 63kg

Time t = 5.0s

Beta energy = 1.5x19^-13J

1 Ci = 3.7x10¹⁰

3.0Ci = 3x3.7x10¹⁰

= 11.1x10¹⁰decay/s

We have a total of 11.1x10¹⁰ beta particles produced per second

The question says there is 100% absorption

Et = (11.1x10¹⁰)x5x1.5x10^-13

= 0.08325

The radiation dose = Et/m

= 0.08325/63 kg

= 0.001321428J/kg

= 0.001321428 grays

Or 1.32142x10^-3 Gy

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Answer:

0.61°

Explanation:

Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.

Pulling force= resistance force

From the formula for pulling force,

F(x)= Fcos(θ)

= 425×cos(35.2)

=347N

The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)

=425×0.567=245N

Resistance force= (325N+ 245N) (α)= 570N(α)

We can now equates the pulling force to resistance force

570 (α)= 347N

(α)= 347/570

= 0.61

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