Answer:
F = 8.6 10⁻¹² N
Explanation:
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Em₀ = U = q ΔV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v²
Em₀ = Emf
e ΔV = ½ m v²
v =√ 2 e ΔV / m
v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)
v = √(1.8075 10¹⁶)
v = 1,344 10⁸ m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Emo = U = q DV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v2
Emo = Emf
.e DV = ½ m v2
.v = RA 2 e DV / m
.v = RA (2 1.6 10-19 51400 / 9.1 10-31)
.v = RA (1.8075 10 16)
.v = 1,344 108 m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10-19 1,344 108 0.4
F = 8.6 10-12 N
I suppose that you wanted write "uncharged". The particles without electrical charge present in the nucleus are called neutrons.
Answer:
i think that it is c because it not only tells us about what the moon does to the earth it also tells us what the earth does to the moon.
Explanation:
Answer:
3.86×10⁶ Newton/coulombs
Explaination:
Applying,
E = F/q....................... Equation 1
Where E = Electric Field, F = Force, q = charge.
From the question,
Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs
Substitute these values into equation 1
E = 5.4×10⁻¹/ -1.4×10⁻⁷
E = -3.86×10⁶ Newtons/coulombs
Hence the magnitude of the electric field created by the
negative test charge is 3.86×10⁶ Newton/coulombs
