Answer:
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Answer:
Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s
Explanation:
Given;
initial speed of proton, u = 2.5 x 10⁵ m/s
initial potential, V = 1500 V
mass of proton = 1.67 x 10⁻²⁷ kg
Work done, W = eV= ΔK.E = ¹/₂mu²
eV = ¹/₂mu² (J)
where;
e is the charge of the proton in coulombs
V is the electric potential in volts
m is the mass of the proton in kg
u is the speed of the proton in m/s


Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s
Answer:
2 N/m²
Explanation:
Pressure is defined as the force acting on a unit area .
Therefore;
Pressure = Force /Area
Force = 4 N
Area = 2 m²
Therefore;
Pressure = 4 N/ 2m²
= 2 N/m²
Answer:
V=16.306m/s
Explanation:
Use V=vi+at
So V= 7.29m/s+(3.22)(2.8s)= 16.306m/s
I am not completely sure, but I believe that it depends on the total mass of the Protons and Neutrons