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
Answer:
d= 794.4 cmExplanation:
Given that
Speed ,V= 286 km/h

V=79.44 m/s
Given that time ,t= 100 ms
t= 0.1 s
We know that ( if acceleration is zero)
Distance = Speed x time
d= V t
Now by putting the values in the above equation
d = 79.44 x 0.1 m
d= 7.944 m
We know that 1 m = 100 cm
d= 794.4 cm
The answers are as follows;
a) the inductive reactance is 322 ohm
b) The maximum voltage is 387.5 V
c) The rms and maximum currents in the inductor are 1.2 A and 0.85 A.
<h3>What is the reactance?</h3>
The reactance is obtained from;
XL = 2πfL
XL = 2 * 3.14 * 57.0 * 0.900
XL = 322 ohm
The maximum voltage is obtained as;
Vo = Vrms * √2
Vo = 274 V * √2
Vo = 387.5 V
Io = Vo/XL
Io = 387.5 V/ 322 ohm
Io = 1.2 A
Irms = 274 V/322 ohm
Irms = 0.85 A
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Use newton's second law, F = ma.
Force is 34 N and a is 4 m/s/s
34 = m(4), solve for m.
34/4 = 8.5 g