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Lina20 [59]
3 years ago
12

At the beginning of an experiment, a scientist has 176 grams of radioactive goo. After 165 minutes, her sample has decayed to 5.

5 grams.
What is the half-life of the goo in minutes?

Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?

How many grams of goo will remain after 50 minutes? ...?
Chemistry
1 answer:
Paul [167]3 years ago
6 0
The half-life equation is written as:

An = Aoe^-kt

We use this equation for the solution. We do as follows:

5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes? 
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE


Find a formula for G(t) , the amount of goo remaining at time t.G(t)=? 

G(t) = 176e^-0.02t

How many grams of goo will remain after 50 minutes? 

G(t) = 176e^-0.02(50) = 64.75 g
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Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  })

Substituting the values into the equation, it becomes

E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  }) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }) = 7.12 × 10⁵(1 - \frac{0.12}{0.1216}) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

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