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13 g —> 0.013 kg
KE = 1/2(m)(v)^2
KE = 1/2(0.013)(8.5)^2
KE = 0.47 J
<span>The electric force is given by:
F = [ k*(q1)*(q2) ] / d^2
F = Electric force
k = Coulomb's constant
q1 = Charge of one proton
q2 = Charge of second proton
d = Distance between centers of mass
Values:
F = unknown
k = 8.98E 9 N-m^2/C^2
q1 = 1.6E-19
q2 = 1.6E-19
d = 1.0E-15 m
Insert values into F = [ k*(q1)*(q2) ] / d^2
F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2
F = </span>229.888 N
answer
the electric force of repulsion between nuclear protons is 229.888 N
Answer:
a) 35.94 ms⁻²
b) 65.85 m
Explanation:
Take down the data:
ρ = 1000kg/m3
a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot, at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:
Ptot = Pgas + Pwater
However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:
Ptot = Pgas
= 6.46 × 10⁵ Pa
The change in pressure is given by the continuity equation:
ΔP = 1/2ρv²
where v is the velocity of the water as it exits the tank.
Calculating:
6.46 × 10⁵ =1/2 ×1000×v²
solving for v, we get v = 35.94 ms⁻²
b) The Bernoulli's equation will be applicable here.
The water is coming out with the same pressure, therefore, the equation will be:
ΔP = ρgh
6.46 × 10⁵ = 1000 x 9.81 x h
h = 65.85 meters
Answer:
The average current is 19.567 A
Solution:
As per the question:
Charge, Q = 
Time, t = 
Now,
We know that current is constituted by the rate of transfer of the charge per unit time. Thus we can write:
I = 
(1)
Now, the charge that was transferred is 86 % of the original value.
Therefore,
We replace Q by 0.86Q in eqn (1):
I = 
