A polonium isotope with an atomic mass of 211.988868 u undergoes alpha decay, resulting in a daughter isotope with an atomic mas
1 answer:
Answer:
K = 9.53 MeV
Explanation:
The kinetic energy that the alpha particle has emitted, is the energy in excess after removing the resting energy of the atoms and the helium nucleus that forms the alpha particle
Since energy and masses are related and cannot be
m₀ c² = c² + m_He c²+ K
K = c² (m₀ - m_{f} - m_He)
the mass of the Helium atom is 4 u
K = (3 10⁸)² (211,988868 -207.976652 - 4,002) 1,661 10⁻²⁷
K = 14,949 10⁻¹¹ (0.0102)
K = 1,527 10⁻¹² J
let's reduce 1 J = 6,242 10¹² MeV
K = 9.53 MeV
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Answer:
h = 0.362 m
Explanation:
The pressure equation with depth is
P₂ = +ρ g h
The gauge pressure is
P2 - = ρ g h
This is the pressure that muscles can create
P₂ - = 3740 Pa
But still the person needs a small pressure for the transfer of gases, so
P₂ - = 3740 - 188 = 3552 Pa
This is the maximum pressure difference, where the person can still breathe,
Let's clear the height
h = 3552 / ρ g
h = 3552 / (1000 9.8)
h = 0.362 m
This is the maximum depth where the person can still breathe normally.
Answer:
a) M = 4,997 10²⁰ kg , b) T = 1.43 10³ s
Explanation:
a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics
v = v₀ - a t
As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction
a = (v₀ - v) / t
a = (15 - (-15)) /9.00 = 30/9
a = 3.33 m / s²
Now we use Newton's second law where force is the force of universal attraction
F = m a
G m M / r² = m a
M = a r² / G
Let's calculate
M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹
M = 4,997 10²⁰ kg
b) The period of the ship's orbit
In this case we have a centripetal acceleration
The radius of the orbit is the radius of the plant plus the height of the ship from the surface
R = + h
R = 1 10⁵ + 2.00 10⁴
R = 12 10⁴ m
F = m a
G m M / R² = m a
Centripetal acceleration is
a = v² / R
The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle
d = 2π R
v = d / t
v = 2π R / T
Let's replace
G m M / R² = m (2π R / T)² / R
G M = R³ 4π² / T²
T² = 4π² R³ / G M
T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)
T² = 6.82 10¹⁶ / 3.33 10¹⁰
T = √ (2,048 10⁶)
T = 1.43 10³ s