Answer:
1.312 x 10⁻¹² J/nucleon
Explanation:
mass of ¹³⁶Ba = 135.905 amu
¹³⁶Ba contain 56 proton and 80 neutron
mass of proton = 1.00728 amu
mass of neutron = 1.00867 amu
mass of ¹³⁶Ba = 56 x 1.00728 amu + 80 x 1.00867 amu
= 137.10128 amu
mass defect = 137.10128 - 135.905
= 1.19628 amu
mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg
= 1.9858 x 10⁻²⁷ Kg
speed of light = 3 x 10⁸ m/s
binding energy,
E = mass defect x c²
E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²
E = 17.87 x 10⁻¹¹ J/atom
now,
binding energy per nucleon =![\dfrac{17.87\times 10^{-11}}{136}](https://tex.z-dn.net/?f=%5Cdfrac%7B17.87%5Ctimes%2010%5E%7B-11%7D%7D%7B136%7D)
= 0.1312 x 10⁻¹¹ J/nucleon
= 1.312 x 10⁻¹² J/nucleon
Answer:
w = 2w₀ the angular velocity of man doubles
Explanation:
In this exercise, releasing the weights reduces the moment of inertia
I= I₀ / 2
Therefore, since the platform system plus man is isolated, the kinetic moment must be conserved
L₀ = L
I₀ w₀ = I w
I₀ w₀ = I₀ / 2 w
w = 2w₀
therefore the angular velocity of man doubles
Answer:
m = 35.98 Kg ≈ 36 Kg
Explanation:
I₀ = 125 kg·m²
R₁ = 1.50 m
ωi = 0.600 rad/s
R₂ = 0.905 m
ωf = 0.800 rad/s
m = ?
We can apply The law of conservation of angular momentum as follows:
Linitial = Lfinal
⇒ Ii*ωi = If*ωf <em>(I)</em>
where
Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m
If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m
Now, we using the equation <em>(I) </em>we have
(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800
⇒ m = 35.98 Kg ≈ 36 Kg
The 1/2 value comes due to the integration of the Newton's law of motion .. like for doing work we do integrate fds vector. when we integrate it .we get it as 1/2mv^2 + constant
.
Both the <span>telescope and the microscope are used to aid the naked eye to see things that a it normally cannot. Both also use lenses to magnify objects.
Have a good day! =)</span>