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3 years ago

A sample of plutonium-239 decays to one-eighth of its original amount after 7.236 x 104 years. What is its half-life?

Physics

1 answer:

vovangra [49]3 years ago

8 0

3010.176 years. 7.236 x 104 = 752.144. 1/2 = 4/8. 4 x 752.144 =
3010.176

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Determine the kinetic energy of a 55kg woman running with the velocity of 5.87m/s

Greeley [361]

The formula for kinetic energy = ½m·v<span>2

1/2 * 55 kg x 5,87 m/s ^2 = 27.5 x </span>34.4569 = <span>947.56475 Joule </span>≈ 948 J

4 0

3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

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3 years ago

ludmilkaskok [199]

Answer:

1) ironing a shirt 2) writing on surfaces 3) working of an eraser

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3 years ago

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klemol [59]

Answer:

A metalloid is used because it is a semiconductor and can become more conductive when more light shines on it

Explanation:

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The most important property they exhibit is that they can become more conductive when more light shines on them. This way more electrons are produced.

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yawa3891 [41]

Answer:

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