Answer:
If energy is conserved, then the sum of the potential energy and the kinetic energy is a constant.
Assuming the proton starts from rest, so it's kineitc energy is zero, but it has a potential energy, PE equal to:
PE = qV
where q =1.6 x 10^-19 C
and V = 1.00 V
Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:
PE* = 0,
KE* = 1/(2mv^2)
Now since
PE + KE = Total energy =PE* + KE*
Therefore,
qV + 0 = 0 + 1/2mv^2
Or
KE = qV = 1.6 10^-19 J
Answer:
aquarium vs the value of both obstacles
Answer:
Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?
Using Coulomb's law:
F = 1/(4πE) x Q1 x Q2/ r^2
Where
k= 1/(4πE) = 9 x 10^9Nm2/C2
Therefore,
F = 9x 10^9 x 2.50 x 10^-5 x2.50 x
10^-5/. ( 0.5)^2
F= 5.625/ 0.25
F= 22.5N approximately
F= 23N.
To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.
Hence To = 23N, attractive. C ans.
Thanks.
