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3 years ago

Can someone help me solve this its a simple series circuit

Physics

1 answer:

daser333 [38]3 years ago

8 0

Answer:

See the explanation below.

Explanation:

Indeed this is a serial circuit, in order to calculate the different questions, we must calculate the current of the circuit. The total resistance of the circuit can be calculated by means of an algebraic summation.

Then using ohm's law,we can calculate the current.

Rt = 2 + 4 + 6 = 12 [ohm]

The current

I = V / Rt

I = 12 / 12

I = 1 [amp]

Therefore the voltage in each resistance will be:

V1 = 2 * 1 = 2 [V]

V2 = 4 * 1 = 4 [V]

V3 = 6 * 1 = 6 [V]

The total voltage will be:

V = V1 + V2 + V3 = 12 [V]

The power can be calculated by multiplying the voltage by the total current.

P = V * I

P = 12 * 1

P = 12 [W]

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A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

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Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

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(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

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We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

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3 years ago

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Answer:

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