The speed of a electron that is accelerated from rest through an electric potential difference of 120 V is 
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How to calculate the speed of the electron?</h3>
We know, that the energy of the system is always conserved.
Using the Law of Conservation of energy,
U=0
Here, K is the kinetic energy and U is the potential energy.
Now, substituting the formula of U and K, we get:
=0------(1)
Here,
m is the mass of the electron
v is the speed of the electron
q is the charge on the electron
V is the potential difference
Let
and
represent the final and initial speed.
Here,
=0
Solving for
, we get:


=6.49
m/s
To learn more about the conservation of energy, refer to:
brainly.com/question/2137260
#SPJ4
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
If I remember correctly (from my studies long time ago) the layers are from the outer to the center:
SiAl : Silicon-Aluminum
SiMa : Silicon-Magnesium (although should be Mg)
NiFe : Nickel-Iron
The SiMa layer should have the lightest elements (Magnesium is lighter than Aluminum)
C- 10ft. Hope this helped. Have a great day! :D
Answer:
hello the diagram related to this question is missing attached below is the missing diagram
Answer :
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south
Explanation:
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south