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3 years ago

Which of the following nuclei is most stable based on its binding energy?

Physics

1 answer:

Anettt [7]3 years ago

4 0

We have that the most stable nuclei are the ones with the highest average binding energy. We see that Nitrogen has a mass number of 15 and that in this region of the graph average binding energy is low. Silver and Gold are along a line where there is a constant decline in average binding energy; silver has more than gold. However, we see that at the start of this decline, there is Fe 56. This region has the elements with the highest average binding energy; Nickel with a mass number of 58 is right there and thus it is the most stable nucleus out of the listed ones.

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The highest point of a wave is called the crest. Among the choices, the correct answer is C. The height of the wave can be determined using the crest and the trough. The trough is the lowest point of a wave. The wavelength is the distance between two crests of a wave.

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2 years ago

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3 years ago

Suppose you need your silicon circuit element to run continuously for 3 minutes before it shuts off long enough to cool back dow

Zigmanuir [339]

The maximum rate at which energy can be added to the circuit element mathematically given as

$MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}$

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Generally, the equation for P is mathematically given as

$P=\ln s \frac{\Delta T}{\Delta t}$

Therefore

$Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}$

$\frac{p}{lnS}=\frac{7.4 \times 10^{-3}}{23 \times 10^{-6} \times 705}$

$\frac{p}{lnS}=0.456^{\circ \mathrm{c}} / \mathrm{sec}$

Max temp Change

$MaxT=5.6^{\circ} \mathrm{C}$

$\text { time }=3 \times 60$

t=180s

In conclusion, Max Energy Rate

$MER =23 \times 10^{-6} \times \frac{301 \times 5.6}{180}$

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1 year ago

A cyclist is coasting at 15 m/s when she starts down a 450 m long slope that is 30 m high. The cyclist and her bicycle have a co

Flura [38]

Answer:

Her speed at the bottom of the slope is 25.665 m/s

Explanation:

Here we have

Initial velocity, v₁= 15 m/s

Final velocity = v₂

The energy balance present in the system can be represented as

$\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = W$

Where:

m = Mass of the cyclist = 70 kg

W = work done by the drag force = $-F_Dd$

Where:

d = Distance traveled = 450 m

Therefore,

$\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = -F_Dd$ and

$v_2^2 =\frac{ \frac{1}{2}mv_1^2 + mgh -F_Dd}{ \frac{1}{2}m} = v_1^2 + 2gh -\frac{ 2F_Dd}{ m} = 15^2 + 2\times 9.8\times 30 - \frac{2\times 12\times 450}{70}$

= 658.714 m²/s²

v₂ = 25.665 m/s

Her speed at the bottom of the slope = 25.665 m/s.

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3 years ago