Answer:
Final Velocity = √(eV/m)
Explanation:
The Workdone, W, in accelerating a charge, 2e, through a potential difference, V is given as a product of the charge and the potential difference
W = (2e) × V = 2eV
And this work is equal to change in kinetic energy
W = Δ(kinetic energy) = ΔK.E
But since the charge starts from rest, initial velocity = 0 and initial kinetic energy = 0
ΔK.E = ½ × (mass) × (final velocity)²
(Velocity)² = (2×ΔK.E)/(mass)
Velocity = √[(2×ΔK.E)/(mass)]
ΔK.E = W = 2eV
mass = 4m
Final Velocity = √[(2×W)/(4m)]
Final Velocity = √[(2×2eV)/4m]
Final Velocity = √(4eV/4m)
Final Velocity = √(eV/m)
Hope this Helps!!!
Answer:
C_{y} = 4.96 and θ' = 104,5º
Explanation:
To add several vectors we can decompose each one of them, perform the sum on each axis, to find the components of the resultant and then find the module and direction.
Let's start by decomposing the two vectors.
Vector A
sin θ =
/ A
cos θ = Aₓ / A
A_{y} = A sin θ
Ax = A cos θ
A_{y} = 4.9 sin 31 = 2.52
Ax = 4.9 cos 31 = 4.20
Vector B
B_{y} = B sin θ
Bx = B cos θ
B_{y} = 6 sin 156 = 2.44
Bx = 6 cos 156 = -5.48
The components of the resulting vector are
X axis
Cx = Ax + B x
Cx = 4.20 -5.48
Cx = -1.28
Axis y
C_{y} = Ay + By
C_{y} = 2.52 + 2.44
C_{y} = 4.96
Let's use the Pythagorean theorem to find modulo
C = √ (Cₙ²x2 + Cy2)
C = Ra (1.28 2 + 4.96 2)
C = 5.12
We use trigonemetry to find the angle
tan θ = C_{y} / Cₓ
θ’ = tan⁻¹ (4.96 / (1.28))
θ’ = 75.5
como el valor de Cy es positivo y Cx es negativo el angulo este en el segundo cuadrante, por lo cual el angulo medido respecto de eje x positivo es
θ’ = 180 – tes
θ‘= 180 – 75,5
θ' = 104,5º
Write the answer below the question?
Answer:
Horse/Speed
55 mph
rounded to the tenth? either 60 or 50
but 350 would stay like that i believe!
Explanation: