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3 years ago

30 points 2 questions HELP!!

Physics

1 answer:

salantis [7]3 years ago

3 0

Answer:

Generally with successive half-life a new element is formed from the decayed nuclei:

isotope - daughter nucleus has same no. of protons (same atomic number) so the same element is formed in the decay

isotone - daughter nucleus has same number of neutrons so the atomic number has changed and a new element is formed (new atomic number)

isobar - daughter nucleus has same mass number - this could be an example of beta decay where the decayed nucleus has the same mass number but loses a neutron and gains a proton because of the lost electron and a new element is formed with the atomic number increases by 1

Example:

U238 Uranium - alpha to Th234

Th234 Thorium - 2 beta to U234 (back to U but atomic mass less by 4)

U234 Uranium - alpha to Th230

Th230 Thorium - alpha to Ra226

etc. but during beta decay (electron loss) a neutron is lost and a proton gained (answer a)

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A tabletop gamer has designed a game that requires three dice to be thrown onto a tray with a measurement grid. To add an extra

Airida [17]

Answer:

The resulting coordinate is $(x,y) = (-0.170 \ m , -0.038 \ m )$

Explanation:

From the question we are told that

The mass of the first dice is $m_1 = 11.10 \ g$

The mass of the second dice is $m_2 = 15.10 \ g$

The mass of the third dice is $m_3 = 18.90 \ g$

The coordinate of the first dice is $(x_1, y_1) = (0.3150\ m , -0.4990 \ m )$

The coordinate of the second dice is $(x_2,y_2) = (-0.4050 \ m , 0.4850 \ m )$

The coordinate of the third dice is $(x_3 , y_3) = (-0.1150 \ m , -0.1850\ m )$

Generally the resulting coordinate of the center of mass of the dice in the x-axis is mathematically evaluated as

$x\ cm = \frac{m_1 * x_1 + m_2 * x_2 + m_3 * x_3}{m_1 + m_2 +m_3 }$

i.e the summation of the moments about their x-axis divided by the magnitude of their masses

substituting values

$x\ cm = \frac{11.0 * 0.3150 + 15.10 * (-0.4050) + 18.90 * (-0.1150)}{11.100 + 15.10 +18.90 }$

$x\ cm = -0.170 \ m$

Generally the resulting coordinate of the center of mass of the dice in the y-axis is mathematically evaluated as

$y\ cm = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3}{m_1 + m_2 +m_3 }$

$y\ cm = \frac{ 11.10 * (-0.4990) + (15.10) * (0.4850) + (18.90) * (-0.1850)}{ 11.10 + 15.10 +18.90 }$

$y\ cm = -0.038 \ m$

Thus the resulting coordinate is $(x,y) = (-0.170 \ m , -0.038 \ m )$

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