Answer:
The neutron loses all of its kinetic energy to nucleus.
Explanation:
Given:
Mass of neutron is 'm' and mass of nucleus is 'm'.
The type of collision is elastic collision.
In elastic collision, there is no loss in kinetic energy of the system. So, total kinetic energy is conserved. Also, the total momentum of the system is conserved.
Here, the nucleus is still. So, its initial kinetic energy is 0. So, the total initial kinetic energy will be equal to kinetic energy of the neutron only.
Now, final kinetic energy of the system will be equal to the initial kinetic energy.
Now, as the nucleus was at rest initially, so the final kinetic energy of the nucleus will be equal to the initial kinetic energy of the neutron.
Thus, all the kinetic energy of the neutron will be transferred to the nucleus and the neutron will come to rest after collision.
Therefore, the neutron loses all of its kinetic energy to nucleus.
Weight = (mass) x (gravity)
Weight = (8 x 10⁻⁴ kg) x (10 N/kg) = 0.008 Newton
Answer:
Explanation:
Given
1 mole of perfect, monoatomic gas
initial Temperature
Work done in iso-thermal process
=initial pressure
=Final Pressure
Since it is a iso-thermal process therefore q=w
Therefore q=39.64 J
(b)if the gas expands by the same amount again isotherm-ally and irreversibly
work done is
Answer:
b) The star is moving away from us.
Explanation:
If an object moves toward us, the light waves it emits are compressed - the wavelength of the light will be shorter, making the light bluer. On the other hand, if an object moves away from us, the light waves are stretched, making it redder. If from laboratory measurements we know that a specific hydrogen spectral line appears at the wavelength of 121.6 nanometers (nm) and the spectrum of a particular star shows the same hydrogen line appearing at the wavelength of 121.8 nm, we can conclude that the star is moving away from npos, since the wavelength related to that star is more expanded.
Answer:
ρ/ρ2 = 3 / R₀ the two densities are different
Explanation:
Density is defined as
ρ = M / V
As the nucleus is spherical
V = 4/3 π r³
Let's replace
ρ = A / (4/3 π R₀³)
ρ = ¾ A / π R₀³
b)
ρ2 = F / area
The area of a sphere is
A = 4π R₀²
ρ2 = F / 4π R₀²
ρ2 = F / 4π R₀²
Atomic number is the number of protons in the nucleon in not very heavy nuclei. This number is equal to the number of neutrons, but changes in heavier nuclei, there are more neutrons than protons.
Let's look for the relationship of the two densities
ρ/ρ2 = ¾ A / π R₀³ / (F / 4π R₀²)
ρ /ρ2 = 3 (A / F) (1 / R₀)
In this case it does not say that the nucleon number is A (F = A), the relationship is
ρ/ρ2 = 3 / R₀
I see that the two densities are different
