Answer:
Explanation:
<u>Conservation of Momentum
</u>
The total momentum of a system of two particles is
Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now
The momentum is conserved if no external forces are acting on the system, thus
Let's put some numbers in the problem and say
120=120
It means that when the particles collide, the first mass returns at 6 m/s and the second continues in the same direction at 28 m/s
Answer:
The temperature of the core raises by 
every second.
Explanation:
Since the average specific heat of the reactor core is 0.3349 kJ/kgC
It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius
Thus reactor core whose mass is 
will require
energy to raise it's temperature by 1 degree Celsius in 1 second
Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by
All of the orbitals in a given subshell have the same value of the "<span>magnetic and principal" quantum number
Hope this helps!</span>
Answer:
a) E = 8628.23 N/C
b) E = 7489.785 N/C
Explanation:
a) Given
R = 5.00 cm = 0.05 m
Q = 3.00 nC = 3*10⁻⁹ C
ε₀ = 8.854*10⁻¹² C²/(N*m²)
r = 4.00 cm = 0.04 m
We can apply the equation
E = Qenc/(ε₀*A) (i)
where
Qenc = (Vr/V)*Q
If Vr = (4/3)*π*r³ and V = (4/3)*π*R³
Vr/V = ((4/3)*π*r³)/((4/3)*π*R³) = r³/R³
then
Qenc = (r³/R³)*Q = ((0.04 m)³/(0.05 m)³)*3*10⁻⁹ C = 1.536*10⁻⁹ C
We get A as follows
A = 4*π*r² = 4*π*(0.04 m)² = 0.02 m²
Using the equation (i)
E = (1.536*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.02 m²)
E = 8628.23 N/C
b) We apply the equation
E = Q/(ε₀*A) (ii)
where
r = 0.06 m
A = 4*π*r² = 4*π*(0.06 m)² = 0.045 m²
Using the equation (ii)
E = (3*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.045 m²)
E = 7489.785 N/C
