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guajiro [1.7K]
3 years ago
8

Phosphorus 32 has a half life of 14.3 days how much of a 4mg sample of phosphorus - 32 will remain after 71.5 days

Physics
1 answer:
andrew-mc [135]3 years ago
6 0

Let us first find out the radioactive constant of Phosphorus 32.

Radioactive constant, λ = \frac{ln2}{t_{1/2}}

Here, {t_{1/2} is half life of phosphorus 32 = 14.3 days

λ = \frac{ln2}{14.3}

The amount of 4 mg of phosphorus remain after 71.5 days can be found using the formula,

m=m₀e⁻(λt)

= 4*e^{-\frac{ln2}{14.3}71.5}

= 4*e^{-ln2*5}

=4*e^{-ln(2)^{5}}

=4*e^{-ln32}

=\frac{4}{32}

= 0.125 mg

The mass of 4 mg of phosphorus 32 remains after 71.5 days will be 0.125 mg.

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Explain the change of frequency of the wave if the tension of the string is increased​
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3 years ago
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
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The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

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3 years ago
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