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Anarel [89]
3 years ago
15

There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.

Physics
1 answer:
Sloan [31]3 years ago
4 0

Via half-life equation we have:


A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }


Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes.  By plugging those values in we get:

A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g


There is 6.25 grams left of Ra-229 after 12 minutes.

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Answer:

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An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1
goblinko [34]

Answer:

v"_{1} = v_{1} tanΘ

v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}Θ

Θ = tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )

Explanation:

Applying the law of conservation of momentum, we have:

Δp_{x = 0}

p_{x} = p"_{x}

m_{1}v_{1} = m_{2}v"_{2} cosΘ (Equation 1)

Δp_{y} = 0

p_{y} = p"_{y}

0 = m_{1} v"_{1} - m_{2} v"_{2} sinΘ (Equation 2)

From Equation 1:

v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}Θ

From Equation 2:

m_{2} v"_{2}sinΘ = m_{1} v_{1}

v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }

Replacing Equation 3 in Equation 4:

v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}

v"_{1}=v_{1}\frac{sinΘ}{cosΘ}

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And we found Θ from the Equation 5:

tanΘ=\frac{v"_{1}}{v_{1}}

Θ=tan^{-1}(\frac{v"_{1}}{v_{1}})

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3 years ago
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