Atomic Mass Unit or the unit used to measure the masses of subatomic particles. This is the number of protons and neutrons. Area outside the atomic nucleus where the probability of finding electrons is high. This is a group of protons and neutrons in the center of all atoms.
Answer:
Explanation:
The mass of the deuteron = mass of the proton + mass of the neutron + mass equivalent of the energy of 2.2 Mev evolved.
I amu = 931 Mev
2.2 Mev = 2.2 / 931 amu
= ( 2.2 / 931 )x 1.6726 x 10⁻²⁷
= .00395 x 10⁻²⁷
The mass of the deuteron =( 1.6726 + 1.6749 + .00395)x 10⁻²⁷ kg
= 3.35145 x 10⁻²⁷ kg
b ) Momentum of gamma ray
= h / λ ( h is plank's constant and λ is wavelength of gamma ray )
= hυ / υλ ( υ is frequency of gamma ray )
= E / c ( E is energy of photon and c is velocity o light )
= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹ J / 3 x 10⁸
= 1.173 x 10⁻²¹ Kg m /s
This will be the momentum of deuteron also
Kinetic energy
= p² / 2m ( p is momentum and m is mass of deuteron )
= ( 1.173 x 10⁻²¹ )² / ( 2 x 3.35145 x 10⁻²⁷)
= 1.376 x ⁻¹⁵ J
Energy of gamma ray
= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹ J
= 3.52 x 10⁻¹³ J
So kinetic energy of deuteron is smaller than energy of gamma ray photon .
Answer:
Machine Efficiency
Explanation:
Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work). The output work is always less than the input work because some of the input work is used to overcome friction. Therefore, efficiency is always less than 100 percent
Answer:
v₂ = 63.62 m / s
Explanation:
For this exercise in fluid mechanics we will use Bernoulli's equation
P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂
where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.
We will assume that the distance between the two parts is small, so y₁ = y₂
P₁-P₂ = ρ g (v₂² - v₁²)
pressure is defined by
P = F / A
we substitute
ΔF / A = ρ g (v₂² - v₁²)
v₂² =
suppose that the area of the wing is A = 1 m²
we substitute
v₂² =
v₂² = 79.10 + 3969
v₂ = √4048.1
v₂ = 63.62 m / s