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harina [27]
2 years ago
15

Uranium-238 (U238) has three more neutrons than uranium-235 (U235). Compared to the speed of sound in a bar of U235, is the spee

d of sound in a bar of U238 higher, lower, or the same? There are several factors that play a role.
(a) Chemically, the atoms of these two isotopes behave in essentially identical ways, since the number of protons (92) is identical and the clouds of 92 electrons are nearly identical. The interatomic distance, and the effective "stiffness" of the interatomic bond, both depend on the chemical properties of the atoms. Therefore, which of the following statements are true?
A) The effective stiffness of the interatomic "spring" is the same in the two isotopes, since this "spring" is a model for the interactions of the outer electrons, which are the same for different isotopes of the same atom.
B) An atom with more mass in the nucleus is bigger, so the distance between neighboring atoms is larger.
C) The interatomic distance is the same for the two isotopes, because it depends on the size of the electron cloud, not the nuclear mass.
D) The effective stiffness of the interatomic "spring" is greater for U238 because the mass is greater.
Physics
1 answer:
natta225 [31]2 years ago
7 0

Answer:

The correct  answers to then question are

A and B

Explanation:

As highlighted, the atoms of the two isotopes are chemically identical and they have similar characteristics due to the presence of equal number of protons  and electrons (92 each). Therefore factors influenced by the chemistry of the atoms of the isotopes such as the distance between atoms and the stiffness of the inter-atomic spring is identical for both  U_{235} and  U_{238}

An analogy of the spring constant and the inter-atomic distance is to consider the atoms of the metal to be interconnected by elastic material and as such the stiffness of the elastic material and the inter-atomic distance are determined by surface factors occurring on the surface

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Answer:

42.99°

Explanation:

F_h = Kinetic friction force

F_{\theta} = Pulling force at angle \theta

N_h = Weight of the box = 150 N

Kinetic friction force

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Pulling force at angle \theta

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According to question

\frac{F_h}{F_{\theta}}=\frac{2}{1}\\\Rightarrow \frac{\muN_h}{\muN_{\theta}}=2\\\Rightarrow \frac{N_h}{N_{\theta}}=2\\\Rightarrow N_{\theta}=\frac{N_h}{2}\\\Rightarrow N_{\theta}=\frac{150}{2}\\\Rightarrow N_{\theta}=75\ N

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