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myrzilka [38]
3 years ago
6

1) A substance has a half life of 20 years. what percentage would be left after 40 years?

Physics
1 answer:
Umnica [9.8K]3 years ago
5 0
1) The half-life is the time required for a substance to reduce to half its initial value. In formulas:
\frac{m(t)}{m_0}  = ( \frac{1}{2} )^{t/t_{1/2}} (1)
where
m(t) is the amount of substance left at time t
m0 is the initial mass
t_{1/2} is the half-life

In this problem, the half-life of the substance is 20 years:
t_{1/2} = 20 y
therefore, the fraction of sample left after t=40 years will be
\frac{m(t)}{m_0}=( \frac{1}{2})^ \frac{40 y}{20 y}  = ( \frac{1}{2})^2 =  \frac{1}{4}

So, only 1/4 of the original sample will be left, which corresponds to 25%.

2) We can use again formula (1), by re-arranging it:
m_0 =  \frac{m(t)} {( \frac{1}{2} )^{ \frac{t}{t_{1/2} }}}
If we use m(t)=10 g (mass of uranium left at time t), and t=4 t_{1/2} (the time is equal to 4 half lifes), we get
m_0 =  \frac{10 g}{ (\frac{1}{2})^4 } =16 \cdot 10 g = 160 g
So, the initial sample of uranium was 160 g.
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A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w
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Answer:

Explanation:

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Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

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When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

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mv² = kA²

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0.12312 = 16•A²

A² = 0.12312/16

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v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

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3 years ago
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