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3 years ago

CALCULATING HALF LIFE

Physics

1 answer:

olga2289 [7]3 years ago

7 0

Answer:

N = 500Bq

Explanation:

No = 2000Bq

Time (T) = 10years

T½ = 5 years.

N = ?

T½ = In2 / λ

T½ = 0.693 / λ

λ = 0.693 / t½

λ = 0.693 / 5

λ = 0.1386

The ratio between the two activities is

(N / No) = e⁻^λt

N = No * e⁻^λt

N = 2000 * e^⁻⁰.¹³⁸⁶ * ¹⁰

N = 2000 * e⁻¹.³⁸⁶

N = 2000 * 0.25

N = 500Bq

The activity of Co-60 after 10 years would be 500Bq

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Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of

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a₀). You know ...
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a₁). Without being told, you assume ...
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3 years ago

A student, who weighs 720N, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is

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Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

Explanation:

The given parameters are;

The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

The acceleration due to gravity is a constant = 9.81 m/s²

Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, $F_{net}$ = The weight of the student + The inertia force of the slowing elevator

∴ The net force, $F_{net}$ = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N

3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

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