A beam of protons enter the electric field of magnitude E = 0.5 N/C between a pair of parallel plates. There is a magnetic field
1 answer:
Answer:
0.217 m/s
Explanation:
The protons in the beam passes undeflected when the electric force is equal to the magnetic force:
qE = qvB
where
q is the proton's charge
E is the magnitude of the electric field
v is the speed of the protons
B is the magnitude of the magnetic field
Re-arranging the equation,
And by substituting
E = 0.5 N/C
B = 2.3 T
We find
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Answer:
F = 2.49 x 10⁻⁹ N
Explanation:
The electrostatic force between two charged bodies is given by Colomb's Law:
where,
F = Electrostatic Force = ?
k = colomb's constant = 9 x 10⁹ N.m²/C²
q₁ = charge on proton = 1.6 x 10⁻¹⁹ C
q₂ = second charge = 1.4 C
r = distace between charges = 0.9 m
Therefore,
<u>F = 2.49 x 10⁻⁹ N</u>
Answer:
sin 2θ = 1 θ=45
Explanation:
They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation
R = Vo² sin 2θ / g
Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.
We calculate the distance traveled for different angle
R = vo² Sin (2 15) /9.8
R = Vo² 0.051 m
In the table are all values in two ways
Angle (θ) distance R (x)
0 0 0
15 0.051 Vo² 0.5 Vo²/g
30 0.088 vo² 0.866 Vo²/g
45 0.102 Vo² 1 Vo²/g
60 0.088 Vo² 0.866 Vo²/g
75 0.051 vo² 0.5 Vo²/g
90 0 0
See graphic ( R Vs θ) in the attached ¡, it can be done with any program, for example EXCEL
