Answer: the correct answer is 7.8026035971 x 10^(-13) joule
Explanation:
Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get
m = 4.00260 u + 222.01757 u = 226.02017 u .
The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is
Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,
which is equivalent to an energy change of
Delta E = (0.00523 u)*(931.5MeV/1u)
Delta E = 4.87 MeV
Converting 4.87 MeV to Joules
1 joule [J] = 6241506363094 mega-electrón voltio [MeV]
4 mega-electrón voltio = 6.40870932 x 10^(-13) joule
4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule
Answer:
B. 4 m/s
Explanation:
v=d/t
Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.
Answer:
The answer is "
"
Explanation:
The formula for velocity:


Answer:
First Question

Second Question
The wavelength is for an X-ray
Explanation:
From the question we are told that
The width of the wall is 
The first excited state is
The ground state is 
Gnerally the energy (in MeV) of the photon emitted when the proton undergoes a transition is mathematically represented as
![E = \frac{h^2 }{ 8 * m * l^2 [ n_1^2 - n_0 ^2 ] }](https://tex.z-dn.net/?f=E%20%20%20%3D%20%20%20%5Cfrac%7Bh%5E2%20%7D%7B%208%20%2A%20m%20%20%2A%20%20l%5E2%20%5B%20n_1%5E2%20-%20n_0%20%5E2%20%5D%20%7D)
Here h is the Planck's constant with value 
m is the mass of proton with value 
So
![E = \frac{( 6.626*10^{-34})^2 }{ 8 * (1.67 *10^{-27}) * (10 *10^{-15})^2 [ 2^2 - 1 ^2 ] }](https://tex.z-dn.net/?f=E%20%20%3D%20%20%20%5Cfrac%7B%28%206.626%2A10%5E%7B-34%7D%29%5E2%20%7D%7B%208%20%2A%20%281.67%20%2A10%5E%7B-27%7D%29%20%20%2A%20%20%2810%20%2A10%5E%7B-15%7D%29%5E2%20%5B%202%5E2%20-%201%20%5E2%20%5D%20%7D)
=> 
Generally the energy of the photon emitted is also mathematically represented as

=> 
=> 
=> 
Generally the range of wavelength of X-ray is 
So this wavelength is for an X-ray.
Answer:
The time rate of change in air density during expiration is 0.01003kg/m³-s
Explanation:
Given that,
Lung total capacity V = 6000mL = 6 × 10⁻³m³
Air density p = 1.225kg/m³
diameter of the trachea is 18mm = 0.018m
Velocity v = 20cm/s = 0.20m/s
dv /dt = -100mL/s (volume rate decrease)
= 10⁻⁴m³/s
Area for trachea =

0 - p × Area for trachea =



⇒

ds/dt = 0.01003kg/m³-s
Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s