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artcher [175]
3 years ago
10

based on the information in the graph why is energy released during the fission of a uranium (U) nucleus?

Physics
2 answers:
frutty [35]3 years ago
4 0

Answer:

B

Explanation:

Nuclear fission is the process of splitting apart nuclei. When nuclei fission energy is released. So much energy is released that there is a measurable decrease in mass, from the mass-energy equivalence. This means that some of the mass is converted to energy.

sasho [114]3 years ago
3 0

Answer:

D

Explanation:

Fission is the break of the nuclei of an unstable atom, releasing heat in the process. By the graph shown, U-238 has less average binding energy per nucleon comparing to U-235. Because it has less binding energy, it is more difficult for it to react (it has less energy to release), so it is more stable than U-235.

The product of its fission is the U-235, an isotope that has a small mass than the uranium. So the mass that is lost during fission is converted to the binding energy in the isotope, making it slightly higher (the difference of mass is only 3 amu).

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Answer:

k = 25.07 N/m

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f = 1.66 Hz

Explanation:

Given:

- The original length of the spring L_o = 50 cm

- The mass hanged m = 230 g

- The amount of stretch given 2x = 18 cm @lowest point.

Find:

a. What is the spring constant? (K=)

b. What is the amplitude of the oscillation?

c. What is the frequency of the oscillation?

Solution:

- Make a FBD of the hanging mass, There are two external forces acting on it that is the force of gravity due to its weight and the springs restoring force when its stretched to its lowest point. After hanging the mass on the spring a new equilibrium position is achieved which also causes the spring to stretch. We can apply the Equilibrium conditions at this point in vertical direction as:

                                      k*x - m*g = 0

                                      k = m*g / x

Where, x is the extension of the spring or mean stretch. which 0.5*amplitude (Lowest point). x = 9 cm

                                      k = 0.23*9.81 / 0.09

                                      k = 25.07 N/m

Answer: For part a we have the stiffness of the spring k = 25.07 N/m

- The amplitude of the oscillating motion is the half the amount of total stretch or the amount the spring extends above or below the mean position.

                                       Amplitude = x = 9 cm

- The frequency of any oscillatory motion which can be modeled by SHM can be expressed as:

                                       f = 1 / 2*p*  sqrt ( k / m )

- Plug the values in:                                        

                                       f = 1 / 2*pi* sqrt (25.07 / 0.23 )

                                       f = 1.66 Hz

Answer: For part c the frequency of oscillation is f = 1.66 Hz

           

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