Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J
Explanation: In order to solve this problem we have to used the second Newton law given by:
∑F= m*a
F-f=m*a where f is the friction force (uk*Normal), from this we have
F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N
then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N
the net Force = (34.5-24.5)N= 10 N
Finally the work done by the net force is equal to kinetic energy change so
W=∫Force net*dr= 10 N* 0.1 m= 1J
Answer:
ρ/ρ2 = 3 / R₀ the two densities are different
Explanation:
Density is defined as
ρ = M / V
As the nucleus is spherical
V = 4/3 π r³
Let's replace
ρ = A / (4/3 π R₀³)
ρ = ¾ A / π R₀³
b)
ρ2 = F / area
The area of a sphere is
A = 4π R₀²
ρ2 = F / 4π R₀²
ρ2 = F / 4π R₀²
Atomic number is the number of protons in the nucleon in not very heavy nuclei. This number is equal to the number of neutrons, but changes in heavier nuclei, there are more neutrons than protons.
Let's look for the relationship of the two densities
ρ/ρ2 = ¾ A / π R₀³ / (F / 4π R₀²)
ρ /ρ2 = 3 (A / F) (1 / R₀)
In this case it does not say that the nucleon number is A (F = A), the relationship is
ρ/ρ2 = 3 / R₀
I see that the two densities are different
Answer:
Q = 12.466μC
Explanation:
For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:

Solving for Q:

Taking special care of all units, we can calculate the value of the charge:
Q = 12.466μC