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weeeeeb [17]
3 years ago
11

Consider the binding energy of two stable nuclei, one with 60 nucleons and one with 200 nucleons. a. Is the total binding energy

of the nucleus with 200 nucleons more than, less than, or equal to the total binding energy of the nucleus with 60 nucleons. Justify your reasoning.
Physics
1 answer:
sashaice [31]3 years ago
4 0

Answer:

The total binding energy of the nucleus with 200 nucleons more than the total binding energy of the nucleus with 60 nucleons

Explanation:

Binding energy can be given by the formula:

Binding energy = Binding energy per nucleon * Number of nucleons

This means that if the binding energy per nucleon = x MeV

Where x is a positive real number

The nucleus with 60 nucleons will have Binding energy = 60x MeV

The nucleus with 200 nucleons will have binding energy = 200x MeV

For a +ve x, 200x > 60x

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tekilochka [14]
Chemical energy (calories) is converted by your body walking on the surface into mechanical/kinetic energy
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3 years ago
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Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
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How many neutrons does element X have if it’s atomic number is 28 and it’s mass number is 81?
labwork [276]

Well let’s put it this way. To find the neutrons you subtract the atomic atomic Nuremberg from the atomic mass. So

Mass=81-Number=28

81-28=53

Final answer is 53.

5 0
3 years ago
What is the angular displacement of a minute hand of a clock after 3 minutes?​
Inessa [10]

Answer:

π/10 rads

Explanation:

It takes an hour (60 minutes) for the minute's hand to turn a full circle or achieve an angular rotation of

2πl rad.

Now, number of periods of 3 minutes in an hour is;

Number of periods = 60/3 = 20 periods

Thus, 3 minutes rotation accounts for 1/20 of 2π the rotation of the minute's hand in an hour.

Thus;

Angular displacement = (1/20) * 2π = π/10 rads

6 0
2 years ago
Can someone help me with this science question
Anettt [7]
Mechanical layers of the earth:

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Chemical layers of the earth:

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Have a great weekend! :)
6 0
3 years ago
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