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3 years ago

A mass of 100 grams of a particular radioactive substance decays according to the function m(t)=100e−t850

Physics

1 answer:

AleksandrR [38]3 years ago

8 0

Answer:

After 1023.4 years the mass of the substance will be 30 g.

Explanation:

Hi there!

Let´s write the function (according to what I found on the web):

$m(t) = 100e^{-t/850}$

We have to find the time "t1" at which the mass of the substance is 30 g. Mathematically:

m(t1) = 30

Then:

$30 = 100e^{-t1/850}$

Let´s solve the equation for t1. First, divide by 100 both sides of the equation:

$0.3 = e^{-t1/850}$ [/tex]

Apply ln to both sides of the equation:

$ln(0.3) = ln(e^{-t/850})$

Use the logarithm property: ln (aᵇ) = b ln(a)

ln(0.3) = -t/850 · ln (e) (ln (e) = 1)

ln(0.3) = -t/850

850 ln(0.3) = -t

t = -850 ln(0.3)

t = 1023.4

After 1023.4 years the mass of the substance will be 30 g.

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Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

$mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=> $w_f = 1.531 \ rad/ s$

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