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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.
Physics
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:

It is given that, the position of a particle as as function of time t is given by :

r(t)=(8t+9)i+(2t^2-8)j+6tk

Let v is the velocity of the particle. Velocity of an object is given by :

v=\dfrac{dr(t)}{dt}

v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}

v=(8i+4tj+6k)\ m/s

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

a=\dfrac{dv(t)}{dt}

a=\dfrac{d[8i+4tj+6k]}{dt}

a=(4j)\ m/s^2

At t = 0, v=(8i+0+6k)\ m/s

v(t)=\sqrt{8^2+6^2} =10\ m/s

Hence, this is the required solution.

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3 years ago
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:

E = 10t^2e^-10t Joules

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Given that the current through a 0.2-H inductor is i(t) = 10te–5t A.

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E = 1/2Ll^2

Substitutes the inductor L and the current I into the formula

E = 1/2 × 0.2 × ( 10te^-5t )^2

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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
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