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CaHeK987 [17]
3 years ago
9

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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Maru [420]

Answer:

q=7.965*10^-^6C

Explanation:

From the question we are told that

Altitude of  d_1 390m

MagnitudeM_1=60.0 N/C

Altitude of d_2=240 m

Magnitude is M_2= 100 N/C

Distance of cube d_c=150 m

Generally the flux \phi is mathematical given as

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Generally Quantity of charge  q is mathematically given as

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3 years ago
A series LRC circuit consists of a 12.0-mH inductor, a 15.0-µF capacitor, a resistor, and a 110-V (rms) ac voltage source. If th
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Answer:

61.85 ohm

Explanation:

L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm

Let ω0 be the resonant frequency.

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\omega _{0}=\frac{1}{\sqrt{12\times 10^{-3}\times 15\times 10^{-6}}}

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Impedance, Z = \sqrt{R^{2}+\left ( XL - Xc \right )^{2}}

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Thus, the impedance at double the resonant frequency is 61.85 ohm.

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A charge of 1.0 × 10-6 μC is located inside a sphere, 1.25 cm from its center. What is the electric flux through the sphere due
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Answer:

\Phi_E=0.11\frac{N\cdot m^2}{C}

Explanation:

According to Gauss's Law, the electric flux of a charged sphere is the electric field multiplied by the area of ​​the spherical surface:

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This is identical to the electric flux of a point charge located in the center of the sphere.

\Phi_E=\frac{1*10^{-12}C}{8.85*10^{-12} \frac{C^2}{N\cdot m^2}}\\\Phi_E=0.11\frac{N\cdot m^2}{C}

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