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stealth61 [152]

3 years ago

5

Francium-223 has a half-life of 22 minutes. A Geiger counter gives a reading of 58 counts per minute (cpm) from a sample of fran

cium-223. What will the reading be 22 minutes later? You can ignore background radiation.

1 answer:

polet [3.4K]3 years ago

8 0

Answer:

29 counts per minute (29 cpm)

Explanation:

The half-life of a radioactive isotope is the time needed for the activity of the sample to halve.The half-life of Francium-223 is exactly 22 minutes: this means that its activity after 22 minutes becomes exactly half of its initial value. Since the initial activity of this sample of francium-223 was 58 cpm, the activity after 22 minutes will be

$A=\frac{58 cpm}{2}=29 cpm$

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Explanation:

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If you are traveling around a curve at a constant speed will your velocity change?

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The current through a 0.2-H inductor is i(t) = 10te–5t A. What is the energy stored in the inductor?

lakkis [162]

Answer:

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3 0

3 years ago

Read 2 more answers

A)If your torsion balance deflects to an angle of 10° when your two spheres are 40cm apart, what angle will it deflect to when t

svp [43]

Answer:

Explanation:

Given

for $\theta=10^{\circ}$

Sphere are $d=40\ cm$

when sphere $d_2=10\ cm$ apart suppose deflection is $\theta _2$

We know

$F=k_t\cdot \theta$

Where F=force between charged particle

$\theta =$ Deflection

$F=\frac{kQ_1Q_2}{r^2}=k_t\cdot \theta$

$\theta =\frac{k}{k_t}\times \frac{Q_1Q_2}{r^2}----1$

thus $\theta \propto \frac{1}{r^2}$

for $\theta _2$

$\frac{\theta _1}{\theta _2}=(\frac{r_2}{r_1})^2$

$\theta _2=16\times \theta _1$

$\theta _2=160^{\circ}$

(b)for $10^{\circ}$ deflection Potential $v_1=8\ kV$

Electric Potential is $V=\frac{kQ}{r}$

$Q=\frac{V\cdot r}{k}$

where V=voltage

k=constant

r=distance between charges

Put value of Q in equation 1

$\theta =\frac{k}{k_t}\times \frac{V^2r^2}{k^2}$

$\theta =\frac{V^2r^2}{k\cdot k_t}$

thus $\theta \propto V^2$

therefore

$\frac{\theta _1}{\theta _2}=(\frac{V_1}{V_2})^2$

$\frac{10}{\theta _2}=(\frac{8}{4})^2$

$\theta _2=\frac{10}{4}=2.5^{\circ}$

5 0

3 years ago