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3 years ago

Which answer shows a balanced nuclear equation for the alpha decay of plutonium-239

1 answer:

7 0

Answer: An alpha particle is a helium-4 nucleus, it contains gwo protons and two neutrons. So, alpha decay, giving off an alpha particle, does the following:

1)reduces the atomic number (number of just protons) by two. Plutonium has atomic number of 94, take two protons away and you are left with 92 which corresponds to uranium.

2) reduces the mass number (protons plus neutrons) by 4. If you start with 239 and subtract two protons and two neutrons, you are left with 235.

Explanation:

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A horse travels 80 meters south in 20 seconds. What is its average velocity?

AURORKA [14]

Average velocity is a vector unit (i.e. includes magnitude <em>and </em>direction) calculated by working out distance ÷ time:

80 metres ÷ 20 seconds = 4 metres/seconds (m/s)

Therefore, your final answer is C. 4 m/s south.

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3 years ago

At what point does the external energy enter the system?

Phoenix [80]

The correct answer as the first one above !

8 0

4 years ago

50 points and brainliest!!!! plz help me

antoniya [11.8K]

decrease yeah free 100 percent

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3 years ago

Read 2 more answers

Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k

Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is $\bf{6.59~m/s}$ .

Part(B): The direction is $\bf{42.7^{0}}$ south to east.

Explanation:

Given:

Mass of Jack, $m_{1} = 59.0~Kg$

Mass of Jill, $m_{2} = 47..0~Kg$

Initial velocity of Jack, $v_{1i} = 8.00~m/s$

Initial velocity of Jill, $v_{2i} = 0$

Final velocity of Jack, $v_{1f} 5.00~m/s$

The final angle made by Jack after collision, $\alpha = 34.0^{0}$

Consider that the final velocity of Jill be $v_{2f}$ and it makes an angle of $\beta$ with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

$~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}$

where, $v_{2f}^{x}$ is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

$v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s$

Conservation of momentum of the system along north direction is given by

$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

where, $v_{2f}^{y}$ is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

$v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s$

Part(B):

The direction is given by

$\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}$

4 0

4 years ago

Freeeeeee pointssssss

Alenkinab [10]

Answer:

Thanks for the points

Explanation:

Points

7 0

3 years ago