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2 years ago

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A large air-filled 0.221-kg plastic ball is thrown up into the air with an initial speed of 11.9 m/s. At a height of 2.77 m, the

ball's speed is 2.85 m/s. What fraction of its original energy has been lost to air friction?

1 answer:

kap26 [50]2 years ago

5 0

Answer:

Initial energy = (M/2)*Vo^2 = 12.80 J

Total energy at 4.7 m and V = 5 m/s:

= (M/2)*5^2 + M*g*4.7 = 1.25 + 4.61 J

Fraction of ball's energy LOST = (12.80-5.86)/12.80

= 0.542

Explanation:

I know it's not the real answer, but this will help you find the real one

Hope this helps!!!

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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

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

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

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

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

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

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

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

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Using formula of activity

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Put the value intyo the formula

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

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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 $2.38\times10^{11}\ nuclei$

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

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Answer:

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The limit for falling off would be $N_{top}=0$ , so the minimum speed would be:

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